Integration By Parts: Definite Integrals
📚 Table of Contents
📐 The Integration by Parts Formula
Integration by Parts is a powerful technique for evaluating integrals of products. It is derived from the product rule for differentiation and allows us to transform difficult integrals into simpler ones. For definite integrals, the formula becomes:
$ \displaystyle \int_{a}^{b} u\,dv \;=\; \bigl[u\,v\bigr]_{a}^{b} \;-\; \int_{a}^{b} v\,du $
This means we first find an antiderivative via integration by parts ($u\,v - \int v\,du$), then evaluate at the limits $(a,b)$. The technique is especially useful for integrals involving products of polynomials with exponential, logarithmic, or trigonometric functions.
Derivation from the Product Rule
The integration by parts formula comes directly from the product rule for derivatives. If we have two differentiable functions $u(x)$ and $v(x)$, then:
$ \frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx} $
Rearranging: $ u \frac{dv}{dx} = \frac{d}{dx}[u \cdot v] - v \frac{du}{dx} $
Integrating both sides: $ \int u\,dv = uv - \int v\,du $
🎯 The LIATE Rule: Choosing u and dv
One of the most critical aspects of integration by parts is choosing which part of the integrand should be $u$ and which should be $dv$. The LIATE rule (also called ILATE) provides a priority ordering for choosing $u$:
L - Logarithmic functions: $\ln(x)$, $\log_b(x)$
I - Inverse trigonometric functions: $\sin^{-1}(x)$, $\tan^{-1}(x)$, $\cos^{-1}(x)$
A - Algebraic functions: $x^n$, polynomials, rational functions
T - Trigonometric functions: $\sin(x)$, $\cos(x)$, $\tan(x)$
E - Exponential functions: $e^x$, $a^x$
The function that appears higher in this list should typically be chosen as $u$, while the function lower in the list becomes $dv$. This ordering is designed so that taking the derivative of $u$ (to get $du$) simplifies the expression, while integrating $dv$ (to get $v$) is manageable.
Why LIATE Works
- Logarithmic and inverse trig functions are difficult to integrate but have simple derivatives
- Polynomials reduce in degree when differentiated, eventually becoming constants
- Exponential and trigonometric functions remain essentially unchanged when differentiated or integrated
Tabular Method (Repeated Integration by Parts)
When integrating products like $x^n e^{ax}$ or $x^n \sin(bx)$ that require multiple applications of integration by parts, the tabular method provides a systematic approach:
- Create two columns: one for successive derivatives of $u$, one for successive integrals of $dv$
- Continue until the $u$ column reaches zero or a pattern emerges
- Multiply diagonally with alternating signs (+, −, +, −, ...)
- Sum all the diagonal products
🔧 Essential Integration Techniques for AP Calculus
When to Use Integration by Parts
Integration by parts is the right choice when you encounter:
- Products of polynomials and exponentials: $\int x^n e^{ax}\,dx$
- Products of polynomials and trigonometric functions: $\int x^n \sin(ax)\,dx$ or $\int x^n \cos(ax)\,dx$
- Products of logarithms and polynomials: $\int x^n \ln(x)\,dx$
- Inverse trigonometric functions: $\int \sin^{-1}(x)\,dx$, $\int \tan^{-1}(x)\,dx$
- Products of exponentials and trigonometric functions (cyclic): $\int e^{ax}\sin(bx)\,dx$
Cyclic Integration by Parts
Sometimes integration by parts leads back to the original integral. This happens with integrals like $\int e^x \sin(x)\,dx$. In such cases:
- Apply integration by parts twice
- You'll get an equation: Original Integral = Expression − (constant) × Original Integral
- Solve algebraically for the original integral
Example: $\int e^x \sin(x)\,dx$
After two applications: $I = e^x \sin(x) - e^x \cos(x) - I$
$2I = e^x(\sin(x) - \cos(x))$
$I = \frac{e^x(\sin(x) - \cos(x))}{2} + C$
Integration by Parts for Definite Integrals
For definite integrals, evaluate the boundary term $[uv]_a^b$ at each step:
$\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du$
where $[uv]_a^b = u(b)v(b) - u(a)v(a)$
📋 Essential Integration Formulas Reference
Before diving into examples, here are the key integration formulas you should know for AP Calculus:
Basic Algebraic Integrals
$ \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) $
$ \int \frac{1}{x}\,dx = \ln|x| + C $
$ \int (ax + b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C $
Exponential Integrals
$ \int e^x\,dx = e^x + C $
$ \int e^{ax}\,dx = \frac{1}{a}e^{ax} + C $
$ \int a^x\,dx = \frac{a^x}{\ln(a)} + C $
$ \int x e^{ax}\,dx = \frac{e^{ax}}{a^2}(ax - 1) + C $
$ \int x^2 e^{ax}\,dx = e^{ax}\left(\frac{x^2}{a} - \frac{2x}{a^2} + \frac{2}{a^3}\right) + C $
Logarithmic Integrals
$ \int \ln(x)\,dx = x\ln(x) - x + C $
$ \int x\ln(x)\,dx = \frac{x^2}{2}\ln(x) - \frac{x^2}{4} + C $
$ \int x^n\ln(x)\,dx = \frac{x^{n+1}}{n+1}\ln(x) - \frac{x^{n+1}}{(n+1)^2} + C $
$ \int [\ln(x)]^2\,dx = x[\ln(x)]^2 - 2x\ln(x) + 2x + C $
Trigonometric Integrals
$ \int \sin(x)\,dx = -\cos(x) + C $
$ \int \cos(x)\,dx = \sin(x) + C $
$ \int \tan(x)\,dx = -\ln|\cos(x)| + C $
$ \int \sec^2(x)\,dx = \tan(x) + C $
$ \int \sin(ax)\,dx = -\frac{1}{a}\cos(ax) + C $
$ \int \cos(ax)\,dx = \frac{1}{a}\sin(ax) + C $
$ \int x\sin(ax)\,dx = \frac{\sin(ax)}{a^2} - \frac{x\cos(ax)}{a} + C $
$ \int x\cos(ax)\,dx = \frac{\cos(ax)}{a^2} + \frac{x\sin(ax)}{a} + C $
Inverse Trigonometric Integrals
$ \int \frac{1}{\sqrt{1-x^2}}\,dx = \sin^{-1}(x) + C $
$ \int \frac{1}{1+x^2}\,dx = \tan^{-1}(x) + C $
$ \int \sin^{-1}(x)\,dx = x\sin^{-1}(x) + \sqrt{1-x^2} + C $
$ \int \tan^{-1}(x)\,dx = x\tan^{-1}(x) - \frac{1}{2}\ln(1+x^2) + C $
Combined Exponential-Trigonometric Integrals
$ \int e^{ax}\sin(bx)\,dx = \frac{e^{ax}}{a^2+b^2}(a\sin(bx) - b\cos(bx)) + C $
$ \int e^{ax}\cos(bx)\,dx = \frac{e^{ax}}{a^2+b^2}(a\cos(bx) + b\sin(bx)) + C $
Reduction Formulas
$ \int x^n e^x\,dx = x^n e^x - n\int x^{n-1}e^x\,dx $
$ \int x^n\sin(x)\,dx = -x^n\cos(x) + n\int x^{n-1}\cos(x)\,dx $
$ \int x^n\cos(x)\,dx = x^n\sin(x) - n\int x^{n-1}\sin(x)\,dx $
$ \int (\ln x)^n\,dx = x(\ln x)^n - n\int (\ln x)^{n-1}\,dx $
📊 Worked Examples: Integration by Parts with Definite Integrals
Below are 20+ examples showing how to apply integration by parts for definite integrals step by step. Each example demonstrates the technique with complete solutions.
Example 1
Evaluate: $ \displaystyle \int_{0}^{1} x e^x \, dx $
Choose $u = x$, hence $du = dx$.
Then $dv = e^x\,dx$ and so $v = e^x$.
Using the formula: $ \int_{0}^{1} x e^x \, dx = \Bigl[x e^x\Bigr]_{0}^{1} - \int_{0}^{1} e^x \, dx $.
Compute separately: $ [x e^x]_{0}^{1} = (1 \cdot e^1) - (0 \cdot e^0) = e $ and $ \int_{0}^{1} e^x\,dx = e - 1$.
So the definite integral is $ e \;-\; (e - 1) = 1$.
$ \displaystyle \int_{0}^{1} x e^x \, dx = 1 $
Example 2
Evaluate: $ \displaystyle \int_{1}^{2} x^2 \sin(x) \, dx $
Take $u = x^2$, so $du = 2x\,dx$.
$dv = \sin(x)\,dx$, thus $v = -\cos(x)$.
$ \int_{1}^{2} x^2 \sin(x)\,dx = \Bigl[ -x^2 \cos(x) \Bigr]_{1}^{2} - \int_{1}^{2} -\cos(x) \cdot 2x \, dx $.
This becomes $ [-x^2 \cos(x)]_{1}^{2} + 2 \int_{1}^{2} x \cos(x)\,dx $.
Evaluate the boundary term: $[-x^2 \cos(x)]_{1}^{2} = \bigl[-(2^2)\cos(2)\bigr] - \bigl[-(1^2)\cos(1)\bigr] = -4\cos(2) + \cos(1)$.
Next, for $ \int x \cos(x)\, dx$, do another integration by parts quickly or recall that $ \int x \cos(x)\, dx = x \sin(x) + \cos(x)$. So $ \int_{1}^{2} x \cos(x)\,dx = \bigl[x \sin(x) + \cos(x)\bigr]_{1}^{2} = \bigl(2\sin(2) + \cos(2)\bigr) - \bigl(1 \cdot \sin(1) + \cos(1)\bigr)$.
Thus the whole integral is: $ -4\cos(2) + \cos(1) \;+\; 2 \bigl[2\sin(2) + \cos(2) - \sin(1) - \cos(1)\bigr] $.
Simplify if desired, but it's perfectly fine in this form.
$ \displaystyle \int_{1}^{2} x^2 \sin(x) \, dx \;=\; -4\cos(2) + \cos(1) \;+\; 2\bigl(2\sin(2) + \cos(2) - \sin(1) - \cos(1)\bigr). $
Example 3
Evaluate: $ \displaystyle \int_{0}^{\pi} x \cos(2x)\, dx $
Let $u = x$ so $du = dx$. Let $dv = \cos(2x)\, dx$. Then $v = \frac{\sin(2x)}{2}$.
Apply integration by parts: $ \int_{0}^{\pi} x \cos(2x)\, dx = \Bigl[x \cdot \tfrac{\sin(2x)}{2}\Bigr]_{0}^{\pi} - \int_{0}^{\pi} \tfrac{\sin(2x)}{2} \, dx $.
Evaluate boundary term: $ \Bigl[x \cdot \tfrac{\sin(2x)}{2}\Bigr]_{0}^{\pi} = \frac{\pi}{2} \sin(2\pi) - 0 \cdot \sin(0) = 0 $.
Remaining integral: $ - \int_{0}^{\pi} \tfrac{\sin(2x)}{2}\, dx = - \tfrac{1}{2} \int_{0}^{\pi} \sin(2x)\, dx. $
Now, $ \int \sin(2x)\, dx = -\tfrac{1}{2}\cos(2x)$. So $ \int_{0}^{\pi} \sin(2x)\, dx = \Bigl[-\tfrac{1}{2}\cos(2x)\Bigr]_{0}^{\pi} = -\tfrac{1}{2}\cos(2\pi) + \tfrac{1}{2}\cos(0) = -\tfrac{1}{2}(1) + \tfrac{1}{2}(1) = 0. $
Therefore the entire integral is $0$.
$ \displaystyle \int_{0}^{\pi} x \cos(2x)\, dx = 0 $
Example 4
Evaluate: $ \displaystyle \int_{2}^{3} x^3 e^x \, dx $
We can do repeated integration by parts or recall a known formula. Let's outline it quickly:
$u = x^3, \quad dv = e^x \, dx$ → $du = 3x^2 \, dx, \quad v = e^x$.
$ \int x^3 e^x\, dx = x^3 e^x - \int 3x^2 e^x\, dx = x^3 e^x - 3 \bigl(\int x^2 e^x\, dx\bigr). $
Similarly, $ \int x^2 e^x\, dx = x^2 e^x - 2 \int x e^x \, dx = x^2 e^x - 2 [ x e^x - e^x ] = x^2 e^x - 2x e^x + 2 e^x. $
Putting it all together yields $ \int x^3 e^x\, dx = x^3 e^x - 3[x^2 e^x - 2x e^x + 2 e^x] = x^3 e^x - 3x^2 e^x + 6x e^x - 6 e^x = (x^3 - 3x^2 + 6x - 6)e^x. $
Therefore, $ \int_{2}^{3} x^3 e^x \, dx = \Bigl[(x^3 - 3x^2 + 6x - 6) e^x\Bigr]_{2}^{3}. $
At $x=3$: $(3^3 - 3 \cdot 3^2 + 6 \cdot 3 - 6) e^3 = (27 - 27 + 18 - 6)e^3 = 12 e^3$.
At $x=2$: $(2^3 - 3 \cdot 2^2 + 6 \cdot 2 - 6) e^2 = (8 - 12 + 12 - 6)e^2 = 2 e^2$.
Subtract: $12 e^3 - 2 e^2$.
$ \displaystyle \int_{2}^{3} x^3 e^x \, dx = 12 e^3 - 2 e^2 $
Example 5
Evaluate: $ \displaystyle \int_{0}^{1} x^2 e^{3x} \, dx $
Let $u = x^2$, so $du = 2x\,dx$. Let $dv = e^{3x}\, dx$. Then $v = \frac{1}{3} e^{3x}$ (because $ \int e^{3x}\, dx = \tfrac{1}{3} e^{3x}$).
Thus $ \int_{0}^{1} x^2 e^{3x}\, dx = \Bigl[ x^2 \cdot \tfrac{1}{3} e^{3x} \Bigr]_{0}^{1} - \int_{0}^{1} \tfrac{1}{3} e^{3x} \cdot 2x \, dx. $
This is $ \Bigl[\tfrac{x^2}{3} e^{3x}\Bigr]_{0}^{1} - \tfrac{2}{3} \int_{0}^{1} x e^{3x}\, dx $.
Evaluate boundary: $ \bigl(\tfrac{1^2}{3} e^{3 \cdot 1}\bigr) - \bigl(\tfrac{0^2}{3} e^{0}\bigr) = \tfrac{1}{3} e^3 $.
Next, $ \int x e^{3x}\, dx$ by parts: let $u = x, \; du = dx$, $dv = e^{3x}dx, \; v = \tfrac{1}{3} e^{3x}$. So $ \int x e^{3x}\, dx = x \cdot \tfrac{1}{3} e^{3x} - \int \tfrac{1}{3} e^{3x}\, dx = \tfrac{x}{3} e^{3x} - \tfrac{1}{3} \cdot \tfrac{1}{3} e^{3x} = \bigl(\tfrac{x}{3} - \tfrac{1}{9}\bigr) e^{3x}. $
Hence $ \int_{0}^{1} x e^{3x} \, dx = \Bigl[\bigl(\tfrac{x}{3} - \tfrac{1}{9}\bigr) e^{3x}\Bigr]_{0}^{1}. $
At $x=1$: $(\tfrac{1}{3} - \tfrac{1}{9}) e^{3} = \tfrac{2}{9} e^3$. At $x=0$: $(0 - 0)\, e^0 = 0$.
So $ \int_{0}^{1} x e^{3x}\, dx = \tfrac{2}{9} e^3$.
Putting it back: $ \int_{0}^{1} x^2 e^{3x}\, dx = \tfrac{1}{3} e^3 - \tfrac{2}{3} \Bigl(\tfrac{2}{9} e^3\Bigr) = \tfrac{1}{3} e^3 - \tfrac{4}{27} e^3 = \Bigl(\tfrac{9}{27} - \tfrac{4}{27}\Bigr) e^3 = \tfrac{5}{27} e^3. $
$ \displaystyle \int_{0}^{1} x^2 e^{3x}\, dx = \frac{5}{27} e^3 $
Example 6
Evaluate: $ \displaystyle \int_{0}^{\frac{\pi}{2}} x \cos(3x)\, dx $
Let $u = x$, thus $du = dx$.
Then $dv = \cos(3x)\, dx$. Integrating, $v = \frac{\sin(3x)}{3}$ because $ \int \cos(3x)\, dx = \tfrac{1}{3}\sin(3x)$.
Using integration by parts: $ \int_{0}^{\frac{\pi}{2}} x \cos(3x)\, dx = \Bigl[x \cdot \frac{\sin(3x)}{3}\Bigr]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \frac{\sin(3x)}{3} \, dx. $
Evaluate the boundary term: $ \Bigl[x \cdot \frac{\sin(3x)}{3}\Bigr]_{0}^{\tfrac{\pi}{2}} = \frac{\pi}{2} \cdot \frac{\sin\bigl(3 \cdot \frac{\pi}{2}\bigr)}{3} - 0 $. Note that $3 \cdot \frac{\pi}{2} = \frac{3\pi}{2}$, and $\sin\bigl(\frac{3\pi}{2}\bigr) = -1$. So this becomes $\frac{\pi}{2} \cdot \frac{-1}{3} = -\frac{\pi}{6}$.
Next, the remaining integral: $ - \int_{0}^{\frac{\pi}{2}} \frac{\sin(3x)}{3}\, dx = -\frac{1}{3} \int_{0}^{\frac{\pi}{2}} \sin(3x)\, dx. $ We have $ \int \sin(3x)\,dx = -\frac{1}{3}\cos(3x)$.
Thus $ \int_{0}^{\frac{\pi}{2}} \sin(3x)\, dx = \Bigl[-\frac{1}{3}\cos(3x)\Bigr]_{0}^{\frac{\pi}{2}} = -\frac{1}{3}\cos\bigl(\tfrac{3\pi}{2}\bigr) - \Bigl(-\frac{1}{3}\cos(0)\Bigr). $ Since $\cos\bigl(\frac{3\pi}{2}\bigr) = 0$, and $\cos(0) = 1$, this becomes $0 + \frac{1}{3} = \frac{1}{3}$.
Hence $ -\frac{1}{3} \int_{0}^{\frac{\pi}{2}} \sin(3x)\, dx = -\frac{1}{3} \cdot \frac{1}{3} = -\frac{1}{9}. $
Combine both parts: $ \Bigl[x \frac{\sin(3x)}{3}\Bigr]_{0}^{\frac{\pi}{2}} + \bigl(-\frac{1}{9}\bigr) = -\frac{\pi}{6} - \frac{1}{9}. $
$ \displaystyle \int_{0}^{\frac{\pi}{2}} x \cos(3x)\, dx = -\frac{\pi}{6} \;-\; \frac{1}{9}. $
Example 7
Evaluate: $ \displaystyle \int_{1}^{3} x^2 e^{2x} \, dx $
Let $u = x^2$, so $du = 2x\,dx$. Let $dv = e^{2x}\, dx$; then $v = \frac{1}{2} e^{2x}$ (since $ \int e^{2x} \, dx = \tfrac{1}{2} e^{2x}$).
By parts, $ \int_{1}^{3} x^2 e^{2x}\, dx = \Bigl[ x^2 \cdot \tfrac{1}{2} e^{2x} \Bigr]_{1}^{3} - \int_{1}^{3} \tfrac{1}{2} e^{2x} \cdot 2x \, dx. $
Simplify: $= \Bigl[\tfrac{x^2}{2} e^{2x}\Bigr]_{1}^{3} - \int_{1}^{3} x e^{2x}\, dx $.
Evaluate boundary term: $ \Bigl[\tfrac{x^2}{2} e^{2x}\Bigr]_{1}^{3} = \Bigl(\tfrac{3^2}{2} e^{6}\Bigr) - \Bigl(\tfrac{1^2}{2} e^{2}\Bigr) = \frac{9}{2} e^{6} - \frac{1}{2} e^{2}. $
Now we handle $ \int x e^{2x}\, dx$ by another integration by parts or known formula: let $u = x$, $du = dx$, $dv = e^{2x}\, dx$, $v = \tfrac{1}{2} e^{2x}$.
So $ \int x e^{2x}\, dx = x \cdot \tfrac{1}{2} e^{2x} - \int \tfrac{1}{2} e^{2x}\, dx = \tfrac{x}{2} e^{2x} - \tfrac{1}{2} \cdot \tfrac{1}{2} e^{2x} = \Bigl(\tfrac{x}{2} - \tfrac{1}{4}\Bigr) e^{2x}. $
Evaluate from 1 to 3: $ \int_{1}^{3} x e^{2x} \, dx = \Bigl[\bigl(\tfrac{x}{2} - \tfrac{1}{4}\bigr) e^{2x}\Bigr]_{1}^{3}. $ At $x=3$: $(\tfrac{3}{2} - \tfrac{1}{4}) e^{6} = \tfrac{5}{4} e^{6}$. At $x=1$: $(\tfrac{1}{2} - \tfrac{1}{4}) e^{2} = \tfrac{1}{4} e^{2}$. So the difference is $\tfrac{5}{4} e^{6} - \tfrac{1}{4} e^{2}$.
Therefore $ \int_{1}^{3} x^2 e^{2x}\, dx = \Bigl(\frac{9}{2} e^{6} - \frac{1}{2} e^{2}\Bigr) - \Bigl(\tfrac{5}{4} e^{6} - \tfrac{1}{4} e^{2}\Bigr). $
Combine like terms: $ \frac{9}{2} e^{6} - \frac{5}{4} e^{6} = \Bigl(\frac{18}{4} - \frac{5}{4}\Bigr)e^{6} = \frac{13}{4} e^{6}. $ and $ -\frac{1}{2} e^{2} + \frac{1}{4} e^{2} = -\frac{2}{4} e^{2} + \frac{1}{4} e^{2} = -\frac{1}{4} e^{2}. $
Thus $ \frac{13}{4} e^{6} \;-\; \frac{1}{4} e^{2} $.
$ \displaystyle \int_{1}^{3} x^2 e^{2x}\, dx = \frac{13}{4} e^{6} \;-\; \frac{1}{4} e^{2}.$
Example 8
Evaluate: $ \displaystyle \int_{-1}^{1} x^3 \sin(2x) \, dx $
Let $u = x^3$, so $du = 3x^2\, dx$. Let $dv = \sin(2x)\, dx$, so $v = -\frac{1}{2}\cos(2x)$ (since $ \int \sin(2x)\,dx = -\frac{1}{2}\cos(2x)$).
Then $ \int_{-1}^{1} x^3 \sin(2x)\, dx = \Bigl[x^3 \bigl(-\tfrac{1}{2}\cos(2x)\bigr)\Bigr]_{-1}^{1} - \int_{-1}^{1} -\tfrac{1}{2} \cos(2x) \cdot 3x^2 \, dx. $
This becomes $ -\tfrac{1}{2} \Bigl[x^3 \cos(2x)\Bigr]_{-1}^{1} + \tfrac{3}{2} \int_{-1}^{1} x^2 \cos(2x)\, dx. $
Evaluate the boundary term: $ \bigl[x^3 \cos(2x)\bigr]_{-1}^{1} = (1^3 \cos(2\cdot1)) - \bigl((-1)^3 \cos(-2)\bigr) = \cos(2) - \bigl(-1 \cdot \cos(-2)\bigr) = \cos(2) + \cos(-2). $ But $\cos(-2) = \cos(2)$, so this is $\cos(2) + \cos(2) = 2\cos(2)$.
Thus $ -\tfrac{1}{2}\cdot 2\cos(2) = -\cos(2). $
Next, $ \int_{-1}^{1} x^2 \cos(2x)\, dx$ is an even function integrand ($x^2$ is even, $\cos(2x)$ is even), so we can integrate from 0 to 1 and double it, or do by parts again. Let's do direct by parts once more for completeness:
Let $u = x^2 \Rightarrow du = 2x\, dx$, $dv = \cos(2x)\, dx \Rightarrow v = \frac{\sin(2x)}{2}$. Then $ \int x^2 \cos(2x)\, dx = x^2 \cdot \frac{\sin(2x)}{2} - \int \frac{\sin(2x)}{2} \cdot 2x \, dx. $
$ = \frac{x^2}{2} \sin(2x) - \int x \sin(2x)\, dx. $ We can do one more parts or recall $ \int x \sin(2x)\, dx$. Let's do it quickly: $ u = x,\; dv = \sin(2x)\,dx \;\Rightarrow\; v = -\tfrac{1}{2}\cos(2x). $ So $ \int x \sin(2x)\,dx = x\bigl(-\tfrac{1}{2}\cos(2x)\bigr) - \int -\tfrac{1}{2} \cos(2x) \, dx = -\tfrac{x}{2}\cos(2x) + \tfrac{1}{2} \int \cos(2x)\,dx = -\tfrac{x}{2}\cos(2x) + \tfrac{1}{2} \cdot \tfrac{1}{2}\sin(2x) = -\tfrac{x}{2}\cos(2x) + \tfrac{1}{4}\sin(2x). $
Therefore, $ \int x^2 \cos(2x)\, dx = \frac{x^2}{2}\sin(2x) - \Bigl[-\tfrac{x}{2}\cos(2x) + \tfrac{1}{4}\sin(2x)\Bigr]. $ $ = \frac{x^2}{2}\sin(2x) + \tfrac{x}{2}\cos(2x) - \tfrac{1}{4}\sin(2x) = \Bigl(\frac{x^2}{2} - \tfrac{1}{4}\Bigr)\sin(2x) + \tfrac{x}{2}\cos(2x). $
We evaluate from -1 to 1: $ F(x) = \Bigl(\frac{x^2}{2} - \tfrac{1}{4}\Bigr)\sin(2x) + \tfrac{x}{2}\cos(2x).$ Notice $x^2/2 - 1/4$ is an even expression, but $\sin(2x)$ is odd, so that product is an odd function. Meanwhile, $(x/2)\cos(2x)$ is an odd times even => odd function. So $F(x)$ is odd, meaning $F(1) = -F(-1)$ but let's just compute directly:
At $x=1$: $\Bigl(\tfrac{1}{2}-\tfrac{1}{4}\Bigr)\sin(2) + \tfrac{1}{2}\cos(2) = \tfrac{1}{4}\sin(2) + \tfrac{1}{2}\cos(2).$
At $x=-1$: $ \Bigl(\tfrac{(-1)^2}{2}-\tfrac{1}{4}\Bigr)\sin(-2) + \tfrac{-1}{2}\cos(-2) = \Bigl(\tfrac{1}{2} - \tfrac{1}{4}\Bigr)(-\sin(2)) + \bigl(-\tfrac{1}{2}\bigr)\cos(2) = \tfrac{1}{4}\cdot(-\sin(2)) - \tfrac{1}{2}\cos(2) = -\tfrac{1}{4}\sin(2) - \tfrac{1}{2}\cos(2). $
Subtracting: $ F(1) - F(-1) = \bigl(\tfrac{1}{4}\sin(2) + \tfrac{1}{2}\cos(2)\bigr) - \bigl(-\tfrac{1}{4}\sin(2) - \tfrac{1}{2}\cos(2)\bigr) = \tfrac{1}{4}\sin(2) + \tfrac{1}{2}\cos(2) + \tfrac{1}{4}\sin(2) + \tfrac{1}{2}\cos(2) = \tfrac{1}{2}\sin(2) + \cos(2). $
Hence $ \int_{-1}^{1} x^2 \cos(2x)\, dx = \tfrac{1}{2}\sin(2) + \cos(2).$
Going back, $ \int_{-1}^{1} x^3 \sin(2x)\, dx = -\cos(2) + \tfrac{3}{2}\bigl(\tfrac{1}{2}\sin(2) + \cos(2)\bigr). $ $ = -\cos(2) + \tfrac{3}{2}\cdot \tfrac{1}{2}\sin(2) + \tfrac{3}{2}\cdot \cos(2) = -\cos(2) + \tfrac{3}{4}\sin(2) + \tfrac{3}{2}\cos(2). $
Combine like terms: $-\cos(2) + \tfrac{3}{2}\cos(2) = \tfrac{1}{2}\cos(2)$.
Final result: $ \tfrac{1}{2}\cos(2) + \tfrac{3}{4}\sin(2). $
$ \displaystyle \int_{-1}^{1} x^3 \sin(2x)\, dx = \frac{1}{2}\cos(2) \;+\; \frac{3}{4}\sin(2).$
Example 9
Evaluate: $ \displaystyle \int_{0}^{2} \bigl(3x - 1\bigr) e^{x} \, dx $
We can distribute or apply integration by parts. Distribution is often simpler: $ \int_{0}^{2} (3x - 1) e^{x}\, dx = 3 \int_{0}^{2} x e^{x}\, dx - \int_{0}^{2} e^{x}\, dx.$
We know $ \int x e^{x}\,dx = (x-1)e^{x}$ (from prior examples). Also $ \int e^{x}\, dx = e^{x}$.
So $ 3 \int_{0}^{2} x e^{x}\, dx = 3 \Bigl[(x-1)e^{x}\Bigr]_{0}^{2} = 3 \Bigl[(2-1)e^{2} - (0-1)e^{0}\Bigr] = 3 \bigl[e^{2} + 1\bigr] = 3e^{2} + 3. $
And $ \int_{0}^{2} e^{x}\, dx = [\,e^{x}\,]_{0}^{2} = e^{2} - 1.$
Hence $ \int_{0}^{2} (3x - 1) e^{x}\, dx = \bigl(3e^{2} + 3\bigr) - (e^{2} - 1) = 3e^{2} + 3 - e^{2} + 1 = 2e^{2} + 4. $
$ \displaystyle \int_{0}^{2} (3x - 1) e^{x} \, dx = 2 e^{2} + 4.$
Example 10
Evaluate: $ \displaystyle \int_{2}^{4} x^{4} e^{x} \, dx $
Similar to earlier repeated integration by parts. We can use the known formula $ \int x^n e^{x}\, dx = e^{x} \cdot P_n(x)$, where $P_n(x)$ is a polynomial of degree $n$. Specifically, we can do it step by step:
$ \int x^4 e^{x}\, dx = x^4 e^{x} - 4 \int x^3 e^{x}\, dx $. Then for $ \int x^3 e^{x}\, dx$, we similarly get $(x^3 - 3x^2 + 6x - 6)e^{x}$ (see Example 4 logic).
Therefore $ \int x^4 e^{x}\, dx = x^4 e^{x} - 4 \bigl[ (x^3 - 3x^2 + 6x - 6)e^{x} \bigr] = x^4 e^{x} - 4x^3 e^{x} + 12x^2 e^{x} - 24x e^{x} + 24 e^{x} $ $ = \bigl(x^4 - 4x^3 + 12x^2 - 24x + 24\bigr)\, e^{x}. $
Hence $ \int_{2}^{4} x^4 e^{x}\, dx = \Bigl[\bigl(x^4 - 4x^3 + 12x^2 - 24x + 24\bigr)\, e^{x}\Bigr]_{2}^{4}. $
At $x=4$: $ (4^4 - 4\cdot4^3 + 12\cdot4^2 - 24\cdot4 + 24) e^{4} = (256 - 256 + 192 - 96 + 24) e^{4} = (120) e^{4}. $
At $x=2$: $ (2^4 - 4\cdot2^3 + 12\cdot2^2 - 24\cdot2 + 24) e^{2} = (16 - 32 + 48 - 48 + 24) e^{2} = 8 e^{2}. $
Subtracting: $120 e^{4} - 8 e^{2}$.
$ \displaystyle \int_{2}^{4} x^4 e^{x}\, dx = 120 e^{4} \;-\; 8 e^{2}.$
Example 11
Evaluate: $ \displaystyle \int_{0}^{\pi} x^3 \cos(2x)\, dx $
Let $u = x^3$ → $du = 3x^2 \, dx$.
Let $dv = \cos(2x)\, dx$ → $v = \tfrac{\sin(2x)}{2}$ (since $ \int \cos(2x)\,dx = \tfrac{1}{2}\sin(2x)$).
Then by parts: $ \int_{0}^{\pi} x^3 \cos(2x)\, dx = \Bigl[x^3 \cdot \tfrac{\sin(2x)}{2}\Bigr]_{0}^{\pi} - \int_{0}^{\pi} \tfrac{\sin(2x)}{2} \cdot 3x^2 \, dx. $
Evaluate boundary term: $ [\, x^3 \cdot \tfrac{\sin(2x)}{2}\, ]_{0}^{\pi} = \tfrac{\pi^3}{2}\sin(2\pi) - 0 = 0 $. (since $ \sin(2\pi) = 0$).
The integral becomes $ - \tfrac{3}{2} \int_{0}^{\pi} x^2 \sin(2x)\, dx. $
Next, we do $ \int x^2 \sin(2x)\, dx$ by parts again or recall a known result. Let $u = x^2$ → $du = 2x\, dx$. $dv = \sin(2x)\, dx$ → $v = -\tfrac{1}{2}\cos(2x)$.
So $ \int x^2 \sin(2x)\, dx = x^2\bigl(-\tfrac{1}{2}\cos(2x)\bigr) - \int -\tfrac{1}{2}\cos(2x)\cdot 2x \, dx = -\tfrac{x^2}{2}\cos(2x) + \int x \cos(2x)\, dx. $
Again, $ \int x \cos(2x)\, dx$ can be done by parts or a known formula. Let $u=x, dv=\cos(2x)\,dx$. Then $du=dx, v=\tfrac{\sin(2x)}{2}$. So $ \int x \cos(2x)\, dx = x \tfrac{\sin(2x)}{2} - \int \tfrac{\sin(2x)}{2}\, dx = \tfrac{x}{2}\sin(2x) - \tfrac{1}{2} \cdot \bigl(-\tfrac{1}{2}\cos(2x)\bigr) = \tfrac{x}{2}\sin(2x) + \tfrac{1}{4}\cos(2x). $
Therefore $ \int x^2 \sin(2x)\, dx = -\tfrac{x^2}{2}\cos(2x) + \bigl(\tfrac{x}{2}\sin(2x) + \tfrac{1}{4}\cos(2x)\bigr) = -\tfrac{x^2}{2}\cos(2x) + \tfrac{x}{2}\sin(2x) + \tfrac{1}{4}\cos(2x). $ $ = \tfrac{x}{2}\sin(2x) + \bigl(-\tfrac{x^2}{2} + \tfrac{1}{4}\bigr)\cos(2x). $
Evaluate from 0 to $\pi$: $F(x) = \tfrac{x}{2}\sin(2x) + \Bigl(-\tfrac{x^2}{2} + \tfrac{1}{4}\Bigr)\cos(2x).$
At $x=\pi$: $ F(\pi) = \tfrac{\pi}{2}\sin(2\pi) + \Bigl(-\tfrac{\pi^2}{2} + \tfrac{1}{4}\Bigr)\cos(2\pi) = 0 + \Bigl(-\tfrac{\pi^2}{2} + \tfrac{1}{4}\Bigr)\cdot 1 = -\tfrac{\pi^2}{2} + \tfrac{1}{4}. $
At $x=0$: $ F(0) = \tfrac{0}{2}\sin(0) + \Bigl(-\tfrac{0^2}{2} + \tfrac{1}{4}\Bigr)\cos(0) = \tfrac{1}{4}\cdot 1 = \tfrac{1}{4}. $
So $ \int_{0}^{\pi} x^2 \sin(2x)\, dx = F(\pi) - F(0) = \bigl(-\tfrac{\pi^2}{2} + \tfrac{1}{4}\bigr) - \tfrac{1}{4} = -\tfrac{\pi^2}{2}. $
Therefore $ \int_{0}^{\pi} x^3 \cos(2x)\, dx = -\tfrac{3}{2} \Bigl(-\tfrac{\pi^2}{2}\Bigr) = \tfrac{3\pi^2}{4}. $
$ \displaystyle \int_{0}^{\pi} x^3 \cos(2x)\, dx = \frac{3\pi^2}{4}. $
Example 12
Evaluate: $ \displaystyle \int_{1}^{2} \bigl(x^2 + 3x\bigr) e^{2x} \, dx $
Distribute: $ \int_{1}^{2} (x^2 + 3x) e^{2x} \, dx = \int_{1}^{2} x^2 e^{2x}\, dx + 3 \int_{1}^{2} x e^{2x}\, dx. $
We can handle each term via known integration by parts. Recall from earlier style: $ \int x e^{ax}\, dx = \frac{x}{a} e^{ax} - \frac{1}{a^2} e^{ax} $ or we do a quick by parts. For $a=2$: $ \int x e^{2x}\, dx = \frac{x}{2} e^{2x} - \int \frac{1}{2} e^{2x}\,dx = \frac{x}{2} e^{2x} - \frac{1}{2}\cdot \frac{1}{2} e^{2x} = \Bigl(\frac{x}{2} - \frac{1}{4}\Bigr)e^{2x}. $
Similarly, $ \int x^2 e^{2x}\, dx $ can be done by parts or recalled: $ \int x^2 e^{2x}\, dx = \frac{x^2}{2} e^{2x} - \int \frac{1}{2} 2x\, e^{2x}\, dx = \frac{x^2}{2} e^{2x} - \int x e^{2x}\, dx. $ Then you plug in the $\int x e^{2x}\, dx$ result above, etc. We only need the definite integral from 1 to 2, so let's do it systematically.
Part (i): $ \int_{1}^{2} x^2 e^{2x}\, dx $. By parts: $ u = x^2,\; dv=e^{2x}dx \;\Rightarrow\; du=2x\,dx,\; v=\tfrac{1}{2} e^{2x}. $ So $ \int x^2 e^{2x}\, dx = x^2 \cdot \tfrac{1}{2} e^{2x} - \int \tfrac{1}{2} e^{2x} \cdot 2x\, dx = \tfrac{x^2}{2} e^{2x} - \int x e^{2x}\, dx. $ Evaluate from 1 to 2, we get $ \Bigl[\tfrac{x^2}{2} e^{2x}\Bigr]_{1}^{2} - \int_{1}^{2} x e^{2x}\, dx. $ The boundary: $ \Bigl[\tfrac{x^2}{2} e^{2x}\Bigr]_{1}^{2} = \Bigl(\tfrac{2^2}{2} e^{4}\Bigr) - \Bigl(\tfrac{1^2}{2} e^{2}\Bigr) = (2 e^{4}) - \tfrac{1}{2} e^{2}. $
Part (ii): $ \int_{1}^{2} x e^{2x}\, dx $. We know $ \int x e^{2x}\, dx = \Bigl(\tfrac{x}{2} - \tfrac{1}{4}\Bigr) e^{2x}. $ So $ \int_{1}^{2} x e^{2x}\, dx = \Bigl[\bigl(\tfrac{x}{2} - \tfrac{1}{4}\bigr) e^{2x}\Bigr]_{1}^{2} = \Bigl(\bigl(\tfrac{2}{2} - \tfrac{1}{4}\bigr) e^{4}\Bigr) - \Bigl(\bigl(\tfrac{1}{2} - \tfrac{1}{4}\bigr) e^{2}\Bigr). $ $ = \Bigl(\tfrac{3}{4} e^{4}\Bigr) - \Bigl(\tfrac{1}{4} e^{2}\Bigr) = \tfrac{3}{4} e^{4} - \tfrac{1}{4} e^{2}. $
Return to part (i): $ \int_{1}^{2} x^2 e^{2x}\, dx = \Bigl(2 e^{4} - \tfrac{1}{2} e^{2}\Bigr) - \Bigl(\tfrac{3}{4} e^{4} - \tfrac{1}{4} e^{2}\Bigr) = 2 e^{4} - \tfrac{1}{2} e^{2} - \tfrac{3}{4} e^{4} + \tfrac{1}{4} e^{2}. $ Combine like terms: $ 2 e^{4} - \tfrac{3}{4} e^{4} = \tfrac{8}{4} e^{4} - \tfrac{3}{4} e^{4} = \tfrac{5}{4} e^{4}. $ $ -\tfrac{1}{2} e^{2} + \tfrac{1}{4} e^{2} = -\tfrac{2}{4} e^{2} + \tfrac{1}{4} e^{2} = -\tfrac{1}{4} e^{2}. $ So $ \int_{1}^{2} x^2 e^{2x}\, dx = \tfrac{5}{4} e^{4} - \tfrac{1}{4} e^{2}. $
Finally, $ \int_{1}^{2} (x^2 + 3x) e^{2x}\, dx = \bigl(\tfrac{5}{4} e^{4} - \tfrac{1}{4} e^{2}\bigr) + 3 \bigl(\tfrac{3}{4} e^{4} - \tfrac{1}{4} e^{2}\bigr) = \tfrac{5}{4} e^{4} - \tfrac{1}{4} e^{2} + \tfrac{9}{4} e^{4} - \tfrac{3}{4} e^{2}. $ $ = \bigl(\tfrac{5}{4} + \tfrac{9}{4}\bigr) e^{4} + \bigl(-\tfrac{1}{4} - \tfrac{3}{4}\bigr) e^{2} = \tfrac{14}{4} e^{4} - \tfrac{4}{4} e^{2} = \tfrac{14}{4} e^{4} - e^{2}. $ $ = \frac{14}{4} e^{4} - e^{2} = \frac{7}{2} e^{4} - e^{2}. $
$ \displaystyle \int_{1}^{2} (x^2 + 3x) e^{2x} \, dx = \frac{7}{2} e^{4} \;-\; e^{2}. $
Example 13
Evaluate: $ \displaystyle \int_{0}^{1} x^2 \ln(1 + x)\, dx $
For integrals with a logarithm, we often choose $u = \ln(1+x)$ so that $du = \frac{1}{1+x} \, dx$, and $dv = x^2 \, dx$. Then $v = \frac{x^3}{3}$.
By parts: $ \int_{0}^{1} x^2 \ln(1+x)\, dx = \Bigl[\ln(1+x)\cdot \tfrac{x^3}{3}\Bigr]_{0}^{1} - \int_{0}^{1} \tfrac{x^3}{3} \cdot \frac{1}{1+x}\, dx. $
Evaluate boundary term: at $x=1$, $\ln(2)\cdot \tfrac{1^3}{3} = \tfrac{\ln(2)}{3}$. at $x=0$, $\ln(1)\cdot \tfrac{0^3}{3}=0$. So that part is $\tfrac{\ln(2)}{3}$.
Now the remaining integral: $ -\int_{0}^{1} \frac{x^3}{3(1+x)}\, dx$, i.e. $ -\tfrac{1}{3} \int_{0}^{1} \frac{x^3}{1+x}\, dx. $
We can simplify $ \frac{x^3}{1+x} = x^2 - x + 1 - \frac{1}{1+x}$ by polynomial division? Let's do it carefully: $ \frac{x^3}{1+x} = x^2 - x + 1 - \frac{1}{1+x}. $ (Because $x^3 = (x^2 - x + 1)(1+x) - 1$.)
So $ \int_{0}^{1} \frac{x^3}{1+x}\, dx = \int_{0}^{1} \Bigl(x^2 - x + 1 - \frac{1}{1+x}\Bigr)\, dx. $
Integrate term by term: $ \int_{0}^{1} x^2\, dx = \Bigl[\tfrac{x^3}{3}\Bigr]_{0}^{1} = \tfrac{1}{3}, $ $ \int_{0}^{1} (-x)\, dx = \Bigl[-\tfrac{x^2}{2}\Bigr]_{0}^{1} = -\tfrac{1}{2}, $ $ \int_{0}^{1} 1\, dx = [\,x\,]_{0}^{1} = 1, $ $ \int_{0}^{1} -\frac{1}{1+x}\, dx = -\Bigl[\ln(1+x)\Bigr]_{0}^{1} = -[\ln(2) - \ln(1)] = -\ln(2). $
Summation: $ \int_{0}^{1} \frac{x^3}{1+x}\, dx = \tfrac{1}{3} - \tfrac{1}{2} + 1 - \ln(2). $ $ = \tfrac{1}{3} + \tfrac{1}{2} + 1 - \tfrac{1}{2} - \ldots $ actually let's be neat: $ \tfrac{1}{3} - \tfrac{1}{2} + 1 - \ln(2) = \Bigl(\tfrac{1}{3} - \tfrac{1}{2}\Bigr) + 1 - \ln(2) = -\tfrac{1}{6} + 1 - \ln(2) = \tfrac{5}{6} - \ln(2). $
Hence $ -\tfrac{1}{3} \int_{0}^{1} \frac{x^3}{1+x}\, dx = -\tfrac{1}{3} \bigl(\tfrac{5}{6} - \ln(2)\bigr) = -\tfrac{5}{18} + \tfrac{1}{3}\ln(2). $
Combine with the boundary term from earlier: $ \int_{0}^{1} x^2 \ln(1+x)\, dx = \tfrac{\ln(2)}{3} + \Bigl(-\tfrac{5}{18} + \tfrac{1}{3}\ln(2)\Bigr) = \tfrac{\ln(2)}{3} + \tfrac{1}{3}\ln(2) - \tfrac{5}{18} = \tfrac{2}{3}\ln(2) - \tfrac{5}{18}. $ $ = \frac{2\ln(2)}{3} - \frac{5}{18}. $
$ \displaystyle \int_{0}^{1} x^2 \ln(1 + x)\, dx = \frac{2}{3}\ln(2) \;-\; \frac{5}{18}. $
Example 14
Evaluate: $ \displaystyle \int_{1}^{2} x^2 \ln(x)\, dx $
Again we use the standard approach with logs: let $u = \ln(x)$, $du = \frac{1}{x}\, dx$. Let $dv = x^2\, dx$, so $v = \tfrac{x^3}{3}$.
By parts: $ \int_{1}^{2} x^2 \ln(x)\, dx = \Bigl[\ln(x)\cdot \tfrac{x^3}{3}\Bigr]_{1}^{2} - \int_{1}^{2} \tfrac{x^3}{3} \cdot \tfrac{1}{x}\, dx. $
Boundary term: $ \Bigl[\tfrac{x^3}{3}\ln(x)\Bigr]_{1}^{2} = \Bigl(\tfrac{2^3}{3}\ln(2)\Bigr) - \Bigl(\tfrac{1^3}{3}\ln(1)\Bigr) = \tfrac{8}{3}\ln(2) - 0. $
Remaining integral: $ - \int_{1}^{2} \tfrac{x^3}{3} \cdot \tfrac{1}{x}\, dx = -\tfrac{1}{3} \int_{1}^{2} x^2 \, dx. $ $ \int_{1}^{2} x^2\, dx = \Bigl[\tfrac{x^3}{3}\Bigr]_{1}^{2} = \tfrac{2^3}{3} - \tfrac{1^3}{3} = \tfrac{8}{3} - \tfrac{1}{3} = \tfrac{7}{3}. $
So $ -\tfrac{1}{3} \cdot \tfrac{7}{3} = -\tfrac{7}{9}. $
Combine: $ \int_{1}^{2} x^2 \ln(x)\, dx = \tfrac{8}{3}\ln(2) - \tfrac{7}{9}. $
$ \displaystyle \int_{1}^{2} x^2 \ln(x)\, dx = \frac{8}{3}\ln(2) \;-\; \frac{7}{9}. $
Example 15
Evaluate: $ \displaystyle \int_{0}^{\frac{\pi}{2}} x^2 \cos^2(x)\, dx $
A common approach for powers of $\cos(x)$ is to use $ \cos^2(x) = \frac{1 + \cos(2x)}{2} $. Then we can handle $ \int x^2 \cos(2x)\, dx$ by parts.
Rewrite: $ \int_{0}^{\frac{\pi}{2}} x^2 \cos^2(x)\, dx = \int_{0}^{\frac{\pi}{2}} x^2 \cdot \frac{1 + \cos(2x)}{2}\, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^2\, dx + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^2 \cos(2x)\, dx. $
The first part is straightforward: $ \frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^2\, dx = \frac{1}{2} \Bigl[\tfrac{x^3}{3}\Bigr]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \cdot \frac{\bigl(\frac{\pi}{2}\bigr)^3}{3} = \frac{1}{2} \cdot \frac{\frac{\pi^3}{8}}{3} = \frac{\pi^3}{48}. $
Next, $ \frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^2 \cos(2x)\, dx. $ Use integration by parts on $ \int x^2 \cos(2x)\, dx$: let $u = x^2$, $dv = \cos(2x)\,dx$. Then $du = 2x\,dx$, $v = \tfrac{\sin(2x)}{2}$.
So $ \int x^2 \cos(2x)\, dx = x^2 \cdot \tfrac{\sin(2x)}{2} - \int \tfrac{\sin(2x)}{2} \cdot 2x\, dx = \tfrac{x^2}{2}\sin(2x) - \int x \sin(2x)\, dx. $
Again, $\int x \sin(2x)\, dx$ by parts or recall known result: let $u = x, dv=\sin(2x)\,dx$, so $du=dx, v=-\tfrac{1}{2}\cos(2x)$. $ \int x \sin(2x)\, dx = x\bigl(-\tfrac{1}{2}\cos(2x)\bigr) - \int -\tfrac{1}{2}\cos(2x)\, dx = -\tfrac{x}{2}\cos(2x) + \tfrac{1}{2}\int \cos(2x)\, dx = -\tfrac{x}{2}\cos(2x) + \tfrac{1}{2}\cdot \tfrac{\sin(2x)}{2} = -\tfrac{x}{2}\cos(2x) + \tfrac{1}{4}\sin(2x). $
Thus $ \int x^2 \cos(2x)\, dx = \tfrac{x^2}{2}\sin(2x) - \Bigl[-\tfrac{x}{2}\cos(2x) + \tfrac{1}{4}\sin(2x)\Bigr] = \tfrac{x^2}{2}\sin(2x) + \tfrac{x}{2}\cos(2x) - \tfrac{1}{4}\sin(2x). $ $ = \bigl(\tfrac{x^2}{2} - \tfrac{1}{4}\bigr)\sin(2x) + \tfrac{x}{2}\cos(2x). $
Evaluate from 0 to $\frac{\pi}{2}$: $F(x) = \bigl(\tfrac{x^2}{2} - \tfrac{1}{4}\bigr)\sin(2x) + \tfrac{x}{2}\cos(2x).$
At $x=\frac{\pi}{2}$: $ F\bigl(\tfrac{\pi}{2}\bigr) = \Bigl(\tfrac{(\pi/2)^2}{2} - \tfrac{1}{4}\Bigr)\sin(\pi) + \tfrac{\pi/2}{2}\cos(\pi). $ Notice $\sin(\pi) = 0$, and $\cos(\pi) = -1$. So that is $ 0 + \Bigl(\tfrac{\pi}{2} \cdot \tfrac{1}{2}\Bigr)\cdot (-1) = \frac{\pi}{4}\cdot(-1) = -\frac{\pi}{4}. $
At $x=0$: $ F(0) = \Bigl(\tfrac{0}{2} - \tfrac{1}{4}\Bigr)\sin(0) + \tfrac{0}{2}\cos(0) = 0. $
Thus $ \int_{0}^{\frac{\pi}{2}} x^2 \cos(2x)\, dx = F\bigl(\tfrac{\pi}{2}\bigr) - F(0) = -\frac{\pi}{4}. $
Finally, $ \int_{0}^{\frac{\pi}{2}} x^2 \cos^2(x)\, dx = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} x^2\, dx + \frac{1}{2}\int_{0}^{\frac{\pi}{2}} x^2 \cos(2x)\, dx $ = $ \frac{\pi^3}{48} + \frac{1}{2}\Bigl(-\frac{\pi}{4}\Bigr) = \frac{\pi^3}{48} - \frac{\pi}{8}. $
$ \displaystyle \int_{0}^{\frac{\pi}{2}} x^2 \cos^2(x)\, dx = \frac{\pi^3}{48} \;-\; \frac{\pi}{8}. $
Example 16
Integral: $ \displaystyle \int_{0}^{1} 2x\, e^{3x}\, dx $
Rewrite as $ 2 \int_{0}^{1} x e^{3x}\, dx$.
Let $u = x,\, dv = e^{3x}dx.$ Then $du=dx,\, v=\tfrac{1}{3}e^{3x}.$
So $ \int x e^{3x}\, dx = x \cdot \tfrac{1}{3} e^{3x} - \int \tfrac{1}{3} e^{3x}\, dx$.
The leftover integral is $\tfrac{1}{3} e^{3x}$, so $ \int x e^{3x}\, dx = \Bigl(\tfrac{x}{3} - \tfrac{1}{9}\Bigr)e^{3x}.$
Evaluate from 0 to 1: $\Bigl(\tfrac{1}{3} - \tfrac{1}{9}\Bigr)e^{3} - \bigl(0 - \tfrac{1}{9}\bigr) = \tfrac{2}{9}e^{3} + \tfrac{1}{9} = \tfrac{2 e^{3} + 1}{9}.$
Multiply by 2: $2 \cdot \tfrac{2 e^{3} + 1}{9} = \tfrac{4 e^{3} + 2}{9}.$
$ \displaystyle \int_{0}^{1} 2x\, e^{3x}\, dx = \frac{4 e^{3} + 2}{9}. $
Example 17
Integral: $ \displaystyle \int_{0}^{\frac{\pi}{2}} x \sin(2x)\, dx $
Let $u = x,\, dv = \sin(2x)\, dx.$ Then $du=dx,\, v= -\tfrac{1}{2}\cos(2x).$
So $ \int x \sin(2x)\, dx = x\bigl(-\tfrac{1}{2}\cos(2x)\bigr) - \int -\tfrac{1}{2}\cos(2x)\, dx.$
Evaluate from 0 to $\tfrac{\pi}{2}$: boundary term at $x=\tfrac{\pi}{2}$ is $(\tfrac{\pi}{2})(-\tfrac{1}{2}\cos(\pi)) = \tfrac{\pi}{4}.$ The leftover integral in cos becomes zero on [0,\tfrac{\pi}{2}].
$ \displaystyle \int_{0}^{\frac{\pi}{2}} x \sin(2x)\, dx = \frac{\pi}{4}. $
Example 18
Integral: $ \displaystyle \int_{1}^{3} x^3 \ln(x)\, dx $
Let $u=\ln(x),\, dv=x^3 dx.$ Then $du=\tfrac{1}{x}dx,\, v=\tfrac{x^4}{4}.$
$ \int x^3 \ln(x)\, dx = \ln(x)\cdot \tfrac{x^4}{4} - \int \tfrac{x^4}{4}\cdot \tfrac{1}{x}\, dx = \tfrac{x^4}{4}\ln(x) - \tfrac{1}{4} \int x^3\, dx.$
$\int x^3\, dx = \tfrac{x^4}{4}.$ So $ \int x^3 \ln(x)\, dx = \tfrac{x^4}{4}\ln(x) - \tfrac{x^4}{16}.$
Evaluate 1 to 3: $ \Bigl[\tfrac{x^4}{4}\ln(x) - \tfrac{x^4}{16}\Bigr]_{1}^{3} = \Bigl(\tfrac{81}{4}\ln(3) - \tfrac{81}{16}\Bigr) - \Bigl(0 - \tfrac{1}{16}\Bigr) = \tfrac{81}{4}\ln(3) - 5.$
$ \displaystyle \int_{1}^{3} x^3 \ln(x)\, dx = \frac{81}{4}\ln(3) \;-\; 5. $
Example 19
Integral: $ \displaystyle \int_{0}^{2} x e^{2x} \, dx $
Let $u=x,\, dv=e^{2x}dx.$ Then $du=dx,\, v=\tfrac{1}{2} e^{2x}.$
$ \int x e^{2x}\, dx = x\bigl(\tfrac{1}{2}e^{2x}\bigr) - \int \tfrac{1}{2} e^{2x}\, dx = \tfrac{x}{2} e^{2x} - \tfrac{1}{4} e^{2x} = \Bigl(\tfrac{x}{2} - \tfrac{1}{4}\Bigr) e^{2x}.$
Evaluate 0 to 2: $ \Bigl(\tfrac{2}{2}-\tfrac{1}{4}\Bigr) e^{4} - \bigl(0-\tfrac{1}{4}\bigr) = \tfrac{3}{4} e^{4} + \tfrac{1}{4}.$
$ \displaystyle \int_{0}^{2} x e^{2x}\, dx = \frac{3}{4} e^{4} + \frac{1}{4}. $
Example 20
Integral: $ \displaystyle \int_{0}^{\frac{\pi}{4}} x^2 \tan(x)\, dx $
Recall $ \int \tan(x)\, dx = -\ln|\cos(x)|.$
Let $u = x^2,\, dv=\tan(x)\,dx.$ Then $du=2x\,dx,\, v=-\ln(\cos x).$
$ \int_{0}^{\tfrac{\pi}{4}} x^2 \tan(x)\, dx = \Bigl[x^2\cdot(-\ln(\cos x))\Bigr]_{0}^{\tfrac{\pi}{4}} - \int_{0}^{\tfrac{\pi}{4}} -\ln(\cos x)\cdot 2x\, dx. $
Boundary at $x=\tfrac{\pi}{4}$ gives $ (\tfrac{\pi}{4})^2 \cdot\bigl(-\ln(\cos(\tfrac{\pi}{4}))\bigr) = -\tfrac{\pi^2}{16}\ln(\tfrac{\sqrt{2}}{2}).$ At $x=0$ is 0.
The rest is $ + 2\int_{0}^{\tfrac{\pi}{4}} x\ln(\cos x)\, dx.$ That integral typically has no simpler closed form, so we leave it there.
$ \displaystyle \int_{0}^{\frac{\pi}{4}} x^2 \tan(x)\, dx = - \frac{\pi^2}{16}\,\ln\!\Bigl(\frac{\sqrt{2}}{2}\Bigr) \;+\; 2 \int_{0}^{\frac{\pi}{4}} x \ln\!\bigl(\cos(x)\bigr)\, dx. $
Summary
We have explored 20+ comprehensive examples applying the definite integral version of the Integration by Parts formula:
$ \displaystyle \int_{a}^{b} u\,dv \;=\; \bigl[u\,v\bigr]_{a}^{b} \;-\; \int_{a}^{b} v\,du $
Each time, we identify parts of the integrand as $u$ and $dv$ using the LIATE rule, then evaluate carefully at the upper and lower limits. This approach allows us to handle integrals of products like $x^n e^x$, $x^n \sin(x)$, $x^n \ln(x)$, and more quite systematically. Remember to practice the tabular method for repeated integration by parts problems.
❓ Frequently Asked Questions (FAQ)
1. What is integration by parts?
Integration by parts is a technique for evaluating integrals of products of two functions. It transforms an integral into a simpler form using the formula $\int u\,dv = uv - \int v\,du$. It's derived from the product rule for differentiation.
2. What is the integration by parts formula for definite integrals?
For definite integrals, the formula is $\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du$, where $[uv]_a^b = u(b)v(b) - u(a)v(a)$. You must evaluate the boundary term at both limits.
3. How do I choose u and dv in integration by parts?
Use the LIATE rule: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Choose $u$ from the function type that appears first in this list. The remaining part becomes $dv$.
4. What is the LIATE rule?
LIATE stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions. It provides an ordering for choosing which function should be $u$ in integration by parts—pick from higher in the list.
5. When should I use integration by parts?
Use integration by parts for products like $x^n e^x$, $x^n \sin(x)$, $x^n \ln(x)$, inverse trig functions, and $e^x \sin(x)$ or $e^x \cos(x)$ (cyclic). If the integral becomes more complex after applying the formula, try a different choice of $u$ and $dv$.
6. What if I need to apply integration by parts multiple times?
For integrals like $\int x^3 e^x\,dx$, you need repeated applications. Use the tabular method: create columns for derivatives of $u$ and integrals of $dv$, multiply diagonally with alternating signs, then sum.
7. What is the tabular method for integration by parts?
The tabular method is a shortcut for repeated integration by parts. Write derivatives of $u$ in one column and integrals of $dv$ in another. Continue until $u$ becomes 0. Multiply diagonally with alternating signs (+, −, +, −, ...).
8. What is cyclic integration by parts?
Cyclic integration occurs when applying integration by parts returns the original integral. This happens with $\int e^x \sin(x)\,dx$. Apply twice, set up an equation with the original integral, and solve algebraically.
9. How do I integrate $\int x e^x\,dx$?
Let $u = x$ and $dv = e^x\,dx$. Then $du = dx$ and $v = e^x$. Apply the formula: $\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1) + C$.
10. How do I integrate $\int \ln(x)\,dx$?
Let $u = \ln(x)$ and $dv = dx$. Then $du = \frac{1}{x}dx$ and $v = x$. Apply the formula: $\int \ln(x)\,dx = x\ln(x) - \int 1\,dx = x\ln(x) - x + C$.
11. How do I integrate products of polynomials and trig functions?
For $\int x^n \sin(ax)\,dx$ or $\int x^n \cos(ax)\,dx$, let $u = x^n$ (algebraic) and $dv$ be the trig part. Apply integration by parts $n$ times until the polynomial disappears. Use the tabular method for efficiency.
12. What are common mistakes in integration by parts?
Common mistakes include: wrong choice of $u$ and $dv$, forgetting the minus sign in the formula, not integrating $dv$ correctly, forgetting boundary terms in definite integrals, and not recognizing cyclic patterns.
13. How does integration by parts relate to the product rule?
Integration by parts is the integral version of the product rule. The product rule states $\frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx}$. Rearranging and integrating gives the integration by parts formula.
14. Can integration by parts be used for inverse trig functions?
Yes! For $\int \sin^{-1}(x)\,dx$ or $\int \tan^{-1}(x)\,dx$, let $u$ be the inverse trig function and $dv = dx$. This converts the difficult integral into a simpler algebraic expression.
15. What is a reduction formula?
A reduction formula expresses an integral in terms of a simpler integral of the same type. For example, $\int x^n e^x\,dx = x^n e^x - n\int x^{n-1}e^x\,dx$. These are derived using integration by parts and are useful for repeated applications.
16. How do I evaluate the boundary term $[uv]_a^b$?
Evaluate $u(x) \cdot v(x)$ at the upper limit $b$ and subtract its value at the lower limit $a$: $[uv]_a^b = u(b)v(b) - u(a)v(a)$. Be careful with trigonometric values at special angles like $0, \pi/2, \pi$, etc.
17. When does integration by parts fail?
Integration by parts "fails" if the resulting integral is more complex than the original. This usually means you chose the wrong $u$ and $dv$. Try switching your choices. Some integrals may require different techniques entirely (substitution, partial fractions, etc.).
18. Is integration by parts on the AP Calculus exam?
Yes! Integration by parts is a core topic in AP Calculus BC. It appears in both multiple choice and free response questions. You should know the formula, LIATE rule, tabular method, and be able to apply it to definite integrals.
19. How do I verify my integration by parts answer?
Differentiate your answer! If you computed $\int f(x)\,dx = F(x) + C$ correctly, then $\frac{d}{dx}[F(x)] = f(x)$. For definite integrals, you can also use numerical methods or a calculator to verify.
20. What are real-world applications of integration by parts?
Integration by parts is used in physics (work, center of mass, moments of inertia), engineering (signal processing, control systems), probability (expected values), and economics (consumer surplus). It's essential for solving differential equations.