Integration by Parts
One of the most important techniques you learn in Calculus II is Integration by Parts. It is used to integrate products of functions (especially when those products do not succumb to a simple substitution). Below is the standard formula:
$ \int u\,dv = u\,v - \int v\,du $
Here:
- $u$ and $dv$ are chosen from the integrand.
- You compute $du$ by differentiating $u$.
- You compute $v$ by integrating $dv$.
Below are 30 worked examples in three difficulty categories: Easy, Medium, and Hard. All final answers have integer coefficients only (no fractions or decimals).
Easy Examples
Example 1 (Easy)
Integral: $ \int x e^x \, dx $
Choose $u = x$ (so $du = dx$) and $dv = e^x \, dx$ (so $v = e^x$).
Apply the formula
$ \int u\,dv = u\,v - \int v\,du $:
$ \int x e^x \, dx = x e^x - \int e^x \, dx $.
The remaining integral is $ \int e^x \, dx = e^x $.
So the result is $ x e^x - e^x + C $.
$ \int x e^x \, dx = (x - 1)\,e^x + C $
Example 2 (Easy)
Integral: $ \int x \cos(x)\, dx $
Let $u = x$, so $du = dx$. Let $dv = \cos(x)\, dx$, so $v = \sin(x)$.
Integration by parts: $ \int x \cos(x)\, dx = x \sin(x) - \int \sin(x)\, dx $.
The remaining integral is $ \int \sin(x)\, dx = -\cos(x) $.
Combine: $ x \sin(x) - ( -\cos(x) ) = x \sin(x) + \cos(x) $.
$ \int x \cos(x)\, dx = x \sin(x) + \cos(x) + C $
Example 3 (Easy)
Integral: $ \int x \sin(x)\, dx $
Take $u = x$, so $du = dx$. Take $dv = \sin(x)\, dx$, so $v = -\cos(x)$.
Then $ \int x \sin(x)\, dx = x(-\cos(x)) - \int -\cos(x)\, dx = -x \cos(x) + \int \cos(x)\, dx $.
$ \int \cos(x)\, dx = \sin(x) $.
So the result is $ -x \cos(x) + \sin(x) $.
$ \int x \sin(x)\, dx = -x \cos(x) + \sin(x) + C $
Example 4 (Easy)
Integral: $ \int x^2 e^x \, dx $
Let $u = x^2$ so $du = 2x\, dx$. Let $dv = e^x \, dx$ so $v = e^x$.
Then $ \int x^2 e^x \, dx = x^2 e^x - \int e^x (2x)\, dx = x^2 e^x - 2 \int x e^x \, dx $.
We already know $ \int x e^x \, dx = (x - 1) e^x $.
Thus the integral becomes $ x^2 e^x - 2[(x - 1) e^x] = x^2 e^x - 2x e^x + 2 e^x = (x^2 - 2x + 2)\, e^x $.
$ \int x^2 e^x \, dx = (x^2 - 2x + 2)\, e^x + C $
Example 5 (Easy)
Integral: $ \int e^x (x + 1)\, dx $
Distribute or do direct integration by parts. Let $u = x+1$ and $dv = e^x\, dx$.
Then $du = dx$, $v = e^x$.
Apply the formula: $ \int (x+1)\,e^x \, dx = (x+1)e^x - \int e^x\, dx $.
$ \int e^x \, dx = e^x $.
So the result is $(x+1)e^x - e^x = x e^x$.
$ \int e^x (x + 1)\, dx = x e^x + C $
Example 6 (Easy)
Integral: $ \int x^3 \, dx $ (using Integration by Parts in a “forced” way!)
While $ \int x^3\, dx $ is simpler by direct power rule, let's artificially apply integration by parts to show how it can still work.
Let $u = x^3$ so $du = 3x^2\, dx$. Let $dv = dx$ so $v = x$.
Then by parts: $ \int x^3\, dx = x^3 \cdot x - \int x \cdot 3x^2\, dx = x^4 - 3 \int x^3\, dx $.
We see $ \int x^3\, dx $ appears on both sides!
Rearrange: $ \int x^3\, dx + 3 \int x^3\, dx = x^4 \Rightarrow 4 \int x^3\, dx = x^4 \Rightarrow \int x^3\, dx = \frac{x^4}{4} $.
$ \int x^3\, dx = \frac{x^4}{4} + C $
Example 7 (Easy)
Integral: $ \int 4x e^x \, dx $
Factor out the 4 if desired: $ \int 4x e^x\, dx = 4 \int x e^x\, dx$.
We know $ \int x e^x\, dx = (x-1)e^x$.
So $ 4 \int x e^x\, dx = 4 \bigl((x - 1)e^x\bigr) = (4x - 4)\,e^x $.
$ \int 4x e^x\, dx = (4x - 4)\, e^x + C $
Example 8 (Easy)
Integral: $ \int (3x)\cos(x)\, dx $
Factor out 3: $ \int 3x \cos(x)\, dx = 3 \int x \cos(x)\, dx$.
From Example 2, $ \int x \cos(x)\, dx = x \sin(x) + \cos(x)$.
Hence $ 3 [ x \sin(x) + \cos(x) ] = 3x \sin(x) + 3 \cos(x)$.
$ \int 3x \cos(x)\, dx = 3x \sin(x) + 3 \cos(x) + C $
Example 9 (Easy)
Integral: $ \int 2x \sin(x)\, dx $
Again factor out 2: $ 2 \int x \sin(x)\, dx$.
From Example 3, $ \int x \sin(x)\, dx = -x \cos(x) + \sin(x)$.
Therefore multiply by 2: $ 2[-x \cos(x) + \sin(x)] = -2x \cos(x) + 2 \sin(x)$.
$ \int 2x \sin(x)\, dx = -2x \cos(x) + 2 \sin(x) + C $
Example 10 (Easy)
Integral: $ \int x^3 e^x \, dx $
We'll do repeated Integration by Parts step by step.
First application: $u = x^3$, $du = 3x^2\, dx$, $dv = e^x\, dx$, $v = e^x$.
$ \int x^3 e^x \, dx = x^3 e^x - \int 3x^2 e^x \, dx$.
Second application: for $ \int x^2 e^x \, dx$, we know from Example 4: $ x^2 e^x - 2x e^x + 2e^x$.
Thus $ \int 3x^2 e^x\, dx = 3 ( x^2 - 2x + 2 ) e^x$.
Combine: $ x^3 e^x - 3[(x^2 - 2x + 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 $.
$ \int x^3 e^x\, dx = (x^3 - 3x^2 + 6x - 6)\, e^x + C $
Medium Examples
Example 11 (Medium)
Integral: $ \int x^2 \sin(x)\, dx $
Let $u = x^2$, so $du = 2x\, dx$. Let $dv = \sin(x)\, dx$, so $v = -\cos(x)$.
Apply the formula: $ x^2 (-\cos(x)) - \int -\cos(x) (2x)\, dx = -x^2 \cos(x) + 2 \int x \cos(x)\, dx $.
We know $ \int x \cos(x)\, dx = x \sin(x) + \cos(x)$.
So $ -x^2 \cos(x) + 2[x \sin(x) + \cos(x)] = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x)$.
$ \int x^2 \sin(x)\, dx = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C $
Example 12 (Medium)
Integral: $ \int x^2 \cos(x)\, dx $
Let $u = x^2$ so $du = 2x\, dx$. Let $dv = \cos(x)\, dx$ so $v = \sin(x)$.
Then $ \int x^2 \cos(x)\, dx = x^2 \sin(x) - \int \sin(x)(2x)\, dx = x^2 \sin(x) - 2 \int x \sin(x)\, dx $.
We have $ \int x \sin(x)\, dx = -x \cos(x) + \sin(x)$.
Substitute: $ x^2 \sin(x) - 2[-x \cos(x) + \sin(x)] = x^2 \sin(x) + 2x \cos(x) - 2 \sin(x)$.
$ \int x^2 \cos(x)\, dx = x^2 \sin(x) + 2x \cos(x) - 2 \sin(x) + C $
Example 13 (Medium)
Integral: $ \int (2x^2 - 3x) e^x \, dx $
Distribute integration: $ \int (2x^2 e^x - 3x e^x)\, dx = 2 \int x^2 e^x \, dx - 3 \int x e^x \, dx $.
Use known results: $ \int x^2 e^x \, dx = (x^2 - 2x + 2)e^x$, $ \int x e^x \, dx = (x - 1)e^x$.
Combine: $ 2[(x^2 - 2x + 2)e^x] - 3[(x - 1)e^x] = (2x^2 - 7x + 7)e^x $.
$ \int (2x^2 - 3x) e^x \, dx = (2x^2 - 7x + 7)\, e^x + C $
Example 14 (Medium)
Integral: $ \int 5x \sin(x)\, dx $
Factor out 5: $ 5 \int x \sin(x)\, dx$.
From earlier, $ \int x \sin(x)\, dx = -x \cos(x) + \sin(x)$.
So $ 5[-x \cos(x) + \sin(x)] = -5x \cos(x) + 5 \sin(x)$.
$ \int 5x \sin(x)\, dx = -5x \cos(x) + 5 \sin(x) + C $
Example 15 (Medium)
Integral: $ \int (x^2 + x) e^x \, dx $
Split: $ \int x^2 e^x \, dx + \int x e^x \, dx$.
Use known formulas: $ \int x^2 e^x \, dx = (x^2 - 2x + 2)e^x$, $ \int x e^x \, dx = (x - 1)e^x$.
Add: $ (x^2 - 2x + 2)e^x + (x - 1)e^x = (x^2 - x + 1)e^x $.
$ \int (x^2 + x) e^x \, dx = (x^2 - x + 1)e^x + C $
Example 16 (Medium)
Integral: $ \int x^3 \cos(x)\, dx $
We will do repeated integration by parts, step by step.
First: $u = x^3$, $du = 3x^2\, dx$, $dv = \cos(x)\, dx$, $v = \sin(x)$.
$ \int x^3 \cos(x)\, dx = x^3 \sin(x) - \int 3x^2 \sin(x)\, dx $.
Second: handle $ \int x^2 \sin(x)\, dx$ from Example 11: $ -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x)$.
Substitute: $ x^3 \sin(x) - 3[-x^2 \cos(x) + 2x \sin(x) + 2 \cos(x)] = x^3 \sin(x) + 3x^2 \cos(x) - 6x \sin(x) - 6 \cos(x)$.
$ \int x^3 \cos(x)\, dx = x^3 \sin(x) + 3x^2 \cos(x) - 6x \sin(x) - 6 \cos(x) + C $
Example 17 (Medium)
Definite Integral: $ \int_{0}^{1} x e^x \, dx $
For definite integrals, Integration by Parts becomes:
Let $u = x$, $du = dx$, $dv = e^x \, dx$, $v = e^x$.
$ \int_{0}^{1} x e^x \, dx = [x e^x]_{0}^{1} - \int_{0}^{1} e^x \, dx $.
Compute: $[x e^x]_{0}^{1} = e - 0 = e$, $ \int_{0}^{1} e^x\, dx = [e^x]_{0}^{1} = e - 1$.
Hence $ e - (e - 1) = 1$.
$ \int_{0}^{1} x e^x \, dx = 1 $
Example 18 (Medium)
Definite Integral: $ \int_{0}^{\pi} x \cos(x)\, dx $
Set $u = x$, $dv = \cos(x)\, dx$.
Then $du = dx$, $v = \sin(x)$.
$ \int_{0}^{\pi} x \cos(x)\, dx = [x \sin(x)]_{0}^{\pi} - \int_{0}^{\pi} \sin(x)\, dx $.
Evaluate: $(\pi \sin(\pi) - 0) = 0$, and $ \int_{0}^{\pi} \sin(x)\, dx = 2$.
Hence $ 0 - 2 = -2$.
$ \int_{0}^{\pi} x \cos(x)\, dx = -2 $
Example 19 (Medium)
Integral: $ \int x^3 \sin(x)\, dx $
Repeated integration by parts: $u = x^3$, $dv = \sin(x)\, dx$, $du = 3x^2\, dx$, $v = -\cos(x)$.
$ \int x^3 \sin(x)\, dx = x^3(-\cos(x)) - \int -\cos(x) \cdot 3x^2\, dx = -x^3 \cos(x) + 3 \int x^2 \cos(x)\, dx$.
Now $ \int x^2 \cos(x)\, dx = x^2 \sin(x) + 2x \cos(x) - 2 \sin(x)$ (Example 12).
Substitute: $ -x^3 \cos(x) + 3[x^2 \sin(x) + 2x \cos(x) - 2 \sin(x)] = -x^3 \cos(x) + 3x^2 \sin(x) + 6x \cos(x) - 6 \sin(x)$.
$ \int x^3 \sin(x)\, dx = -x^3 \cos(x) + 3x^2 \sin(x) + 6x \cos(x) - 6 \sin(x) + C $
Example 20 (Medium)
Definite Integral: $ \int_{-1}^{1} x \sin(x)\, dx $
Let $u = x$, $dv = \sin(x)\, dx$, so $du = dx$, $v = -\cos(x)$.
$ \int_{-1}^{1} x \sin(x)\, dx = [-x \cos(x)]_{-1}^{1} - \int_{-1}^{1} -\cos(x)\, dx = [-x \cos(x)]_{-1}^{1} + \int_{-1}^{1} \cos(x)\, dx $.
Evaluate each part: $ [-x \cos(x)]_{-1}^{1} = -\cos(1) - (+\cos(1)) = -2\cos(1)$, $ \int_{-1}^{1} \cos(x)\, dx = 2 \sin(1)$.
Total: $ -2 \cos(1) + 2 \sin(1)$.
$ \int_{-1}^{1} x \sin(x)\, dx = -2 \cos(1) + 2 \sin(1) $
Hard Examples
Example 21 (Hard)
Integral: $ \int x^4 e^x \, dx $
We apply integration by parts multiple times.
First: $u = x^4$, $du = 4x^3\, dx$, $dv = e^x\, dx$, $v = e^x$.
$ \int x^4 e^x \, dx = x^4 e^x - \int 4x^3 e^x\, dx$.
Second: from Example 10, $ \int x^3 e^x\, dx = (x^3 - 3x^2 + 6x - 6)e^x$.
Combine: $ x^4 e^x - 4[(x^3 - 3x^2 + 6x - 6)e^x] = x^4 e^x - 4x^3 e^x + 12x^2 e^x - 24x e^x + 24 e^x = (x^4 - 4x^3 + 12x^2 - 24x + 24)\, e^x$.
$ \int x^4 e^x \, dx = (x^4 - 4x^3 + 12x^2 - 24x + 24)\, e^x + C $
Example 22 (Hard)
Integral: $ \int_{0}^{\pi} x^3 \cos(x)\, dx $
We can use the indefinite result from Example 16 and evaluate at 0 and $ \pi$: $ x^3 \sin(x) + 3x^2 \cos(x) - 6x \sin(x) - 6 \cos(x)$.
Evaluate: $ [x^3 \sin(x) + 3x^2 \cos(x) - 6x \sin(x) - 6 \cos(x)]_{0}^{\pi}$.
At $x = \pi$: $ \pi^3 \cdot 0 + 3(\pi^2)(-1) - 6(\pi)(0) - 6(-1) = -3\pi^2 + 6$.
At $x = 0$: $ 0 + 0 - 0 - 6(1) = -6$.
Difference: $ [-3\pi^2 + 6] - [-6] = -3\pi^2 + 12$.
$ \int_{0}^{\pi} x^3 \cos(x)\, dx = -3\pi^2 + 12 $
Example 23 (Hard)
Integral: $ \int (x^4 - x) e^x \, dx $
We can separate or do repeated integration by parts in one shot.
Split: $ \int x^4 e^x \, dx - \int x e^x \, dx$.
We know $ \int x^4 e^x \, dx = (x^4 - 4x^3 + 12x^2 - 24x + 24)e^x$ and $ \int x e^x \, dx = (x - 1)e^x$.
Combine: $ [(x^4 - 4x^3 + 12x^2 - 24x + 24) e^x] - [(x - 1)e^x] = (x^4 - 4x^3 + 12x^2 - 24x + 24 - x + 1)e^x = (x^4 - 4x^3 + 12x^2 - 25x + 25)e^x$.
$ \int (x^4 - x) e^x \, dx = (x^4 - 4x^3 + 12x^2 - 25x + 25)\, e^x + C $
Example 24 (Hard)
Integral: $ \int_{0}^{\frac{\pi}{2}} x^2 \sin(x)\, dx $
From Example 11 (indefinite): $ \int x^2 \sin(x)\, dx = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x)$.
Evaluate at $ x = \frac{\pi}{2}$: $ -\Bigl(\frac{\pi}{2}\Bigr)^2 \cdot 0 + 2\Bigl(\frac{\pi}{2}\Bigr)\cdot 1 + 2(0) = \pi $.
At $ x = 0$: $ -(0^2)(1) + 2(0)(0) + 2(1) = 2 $.
Difference: $ \pi - 2$.
$ \int_{0}^{\frac{\pi}{2}} x^2 \sin(x)\, dx = \pi - 2 $
Example 25 (Hard)
Integral: $ \int (x^3 + 2x^2) e^x \, dx $
Split: $ \int x^3 e^x \, dx + 2 \int x^2 e^x \, dx$.
Known results: $ \int x^3 e^x \, dx = (x^3 - 3x^2 + 6x - 6)e^x$, $ \int x^2 e^x \, dx = (x^2 - 2x + 2)e^x$.
Combine: $ (x^3 - 3x^2 + 6x - 6)e^x + 2[(x^2 - 2x + 2)e^x] = (x^3 - x^2 + 2x - 2)e^x$.
$ \int (x^3 + 2x^2) e^x \, dx = (x^3 - x^2 + 2x - 2)\, e^x + C $
Example 26 (Hard)
Integral: $ \int x^4 \sin(x)\, dx $
Another longer repeated-by-parts.
$u = x^4$, $dv = \sin(x)\, dx$, $du = 4x^3\, dx$, $v = -\cos(x)$.
$ \int x^4 \sin(x)\, dx = - x^4 \cos(x) + 4 \int x^3 \cos(x)\, dx $.
We have $ \int x^3 \cos(x)\, dx = x^3 \sin(x) + 3x^2 \cos(x) - 6x \sin(x) - 6 \cos(x)$ (Example 16).
Multiply by 4: $ - x^4 \cos(x) + 4[x^3 \sin(x) + 3x^2 \cos(x) - 6x \sin(x) - 6 \cos(x)] = - x^4 \cos(x) + 4x^3 \sin(x) + 12x^2 \cos(x) - 24x \sin(x) - 24 \cos(x)$.
$ \int x^4 \sin(x)\, dx = - x^4 \cos(x) + 4x^3 \sin(x) + 12x^2 \cos(x) - 24x \sin(x) - 24 \cos(x) + C $
Example 27 (Hard)
Integral: $ \int x^5 e^x \, dx $
One more repeated integration by parts with polynomials times $e^x$.
Generally, $ \int x^n e^x \, dx$ yields a polynomial of degree $n$ times $e^x$.
Let $u = x^5$, $dv = e^x\, dx$, so $du = 5x^4\, dx$, $v = e^x$.
$ \int x^5 e^x \, dx = x^5 e^x - 5 \int x^4 e^x \, dx$.
Use Example 21 for $ \int x^4 e^x \, dx$: $(x^4 - 4x^3 + 12x^2 - 24x + 24)e^x$.
Combine: $ x^5 e^x - 5[(x^4 - 4x^3 + 12x^2 - 24x + 24)e^x] = (x^5 - 5x^4 + 20x^3 - 60x^2 + 120x - 120)\, e^x $.
$ \int x^5 e^x \, dx = (x^5 - 5x^4 + 20x^3 - 60x^2 + 120x - 120)\, e^x + C $
Example 28 (Hard)
Integral: $ \int (2x^3 - 3x^2 + x) e^x \, dx $
Break into separate integrals: $ 2 \int x^3 e^x \, dx - 3 \int x^2 e^x \, dx + \int x e^x \, dx$.
Use known results: $ \int x^3 e^x \, dx = (x^3 - 3x^2 + 6x - 6)e^x$, $ \int x^2 e^x \, dx = (x^2 - 2x + 2)e^x$, $ \int x e^x \, dx = (x - 1)e^x$.
Combine carefully: $ 2[(x^3 - 3x^2 + 6x - 6)e^x] - 3[(x^2 - 2x + 2)e^x] + [(x - 1)e^x]. $
Simplify inside parentheses: $2x^3 - 6x^2 + 12x - 12 - 3x^2 + 6x - 6 + x - 1 = 2x^3 - 9x^2 + 18x - 19$.
$ \int (2x^3 - 3x^2 + x) e^x \, dx = (2x^3 - 9x^2 + 18x - 19)\, e^x + C $
Example 29 (Hard)
Integral: $ \int 3x^4 \cos(x)\, dx $
Factor out 3 and use $ \int x^4 \cos(x)\, dx$, or do repeated parts.
$3 \int x^4 \cos(x)\, dx$.
$ \int x^4 \cos(x)\, dx = x^4 \sin(x) - \int 4x^3 \sin(x)\, dx$.
We know $ \int x^3 \sin(x)\, dx = -x^3 \cos(x) + 3x^2 \sin(x) + 6x \cos(x) - 6 \sin(x)$.
Hence $ -4 \int x^3 \sin(x)\, dx = 4x^3 \cos(x) - 12x^2 \sin(x) - 24x \cos(x) + 24 \sin(x). $
Sum: $ x^4 \sin(x) + 4x^3 \cos(x) - 12x^2 \sin(x) - 24x \cos(x) + 24 \sin(x)$.
Multiply by 3: $ 3[x^4 \sin(x) + 4x^3 \cos(x) - 12x^2 \sin(x) - 24x \cos(x) + 24 \sin(x)]. $
$ \int 3x^4 \cos(x)\, dx = 3x^4 \sin(x) + 12x^3 \cos(x) - 36x^2 \sin(x) - 72x \cos(x) + 72 \sin(x) + C $
Example 30 (Hard)
Definite Integral: $ \int_{0}^{2} x^2 e^x \, dx $
From Example 4 or 11: $ \int x^2 e^x \, dx = (x^2 - 2x + 2) e^x$.
Evaluate: $ [(x^2 - 2x + 2)e^x]_{0}^{2} $.
At $ x=2$: $(2^2 - 4 + 2)e^2 = 2 e^2$.
At $ x=0$: $(0 - 0 + 2)e^0 = 2$.
Difference: $ 2e^2 - 2$.
$ \int_{0}^{2} x^2 e^x \, dx = 2 e^2 - 2 $
Summary
We have demonstrated how to use Integration by Parts on a variety of examples, including polynomials times exponentials, polynomials times trigonometric functions, and definite integrals. The key formula is always
When powers of $x$ are involved, you repeatedly reduce the power. When integrating trig functions multiplied by polynomials, you likewise reduce the polynomial degree with each application. Eventually you get an integral you know, or the same integral reappears and you can solve for it.
