Integrals Involving Trig Functions
In Calculus II, you often encounter integrals that involve various powers of sine, cosine, tangent, or other trigonometric functions. There are several techniques to handle these, including rewriting with trig identities, using substitutions, or applying integration by parts. Below are some examples with detailed steps.
Example 1: $\int \sin^5(x)\,dx$
We want to integrate $ \sin^5(x) $. One strategy for odd powers of sine is to separate out a single $\sin(x)$ and convert the rest into cosines via $\sin^2(x) = 1 - \cos^2(x)$.
Write $\sin^5(x) = \sin^4(x)\,\sin(x)$.
Note that $\sin^4(x) = (\sin^2(x))^2 = (1 - \cos^2(x))^2$. So $\sin^5(x) = (1 - \cos^2(x))^2 \sin(x)$.
Now let $u = \cos(x)$. Then $\frac{du}{dx} = -\sin(x)$, or $du = -\sin(x)\,dx$. Hence $ \sin(x)\,dx = -du$.
The integral becomes: $ \int \sin^5(x)\,dx = \int (1 - \cos^2(x))^2 \sin(x)\,dx = \int (1 - u^2)^2 \cdot (-du). $
So: $ \int \sin^5(x)\,dx = -\int (1 - u^2)^2 \,du = -\int (1 - 2u^2 + u^4)\,du. $
Integrate term by term: $ \int (1 - 2u^2 + u^4)\, du = \int 1\,du - 2\int u^2\,du + \int u^4\,du = u - \tfrac{2u^3}{3} + \tfrac{u^5}{5}. $
Don’t forget the negative sign outside: $ -\bigl[u - \tfrac{2u^3}{3} + \tfrac{u^5}{5}\bigr] = -u + \tfrac{2u^3}{3} - \tfrac{u^5}{5}. $
Finally, substitute back $u = \cos(x)$: $ -\cos(x) + \tfrac{2\cos^3(x)}{3} - \tfrac{\cos^5(x)}{5} + C. $
$ \displaystyle \int \sin^5(x)\,dx = -\cos(x) + \frac{2\cos^3(x)}{3} - \frac{\cos^5(x)}{5} + C. $
Example 2: $\int \cos(x)\,\sin^5(x)\,dx$
Now we have $\cos(x)\sin^5(x)$. This time, the presence of $\cos(x)$ is helpful for a direct substitution of $\sin(x)$.
Let $u = \sin(x)$. Then $\frac{du}{dx} = \cos(x)$, so $du = \cos(x)\,dx$.
The integral becomes: $ \int \cos(x)\sin^5(x)\,dx = \int u^5 \,du = \frac{u^6}{6} + C. $
Substitute back $u=\sin(x)$: $ \frac{\sin^6(x)}{6} + C. $
$ \int \cos(x)\,\sin^5(x)\,dx = \frac{\sin^6(x)}{6} + C. $
Example 3: $\int \sin^2(x)\,dx$
A common approach for integrals of $\sin^2(x)$ or $\cos^2(x)$ is to use the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$.
Rewrite: $ \sin^2(x) = \frac{1 - \cos(2x)}{2}. $
Thus: $ \int \sin^2(x)\, dx = \int \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2}\int 1\,dx - \frac{1}{2}\int \cos(2x)\,dx. $
$\int 1\, dx = x$. Also $ \int \cos(2x)\,dx = \frac{\sin(2x)}{2}$.
Hence: $ \int \sin^2(x)\, dx = \frac{x}{2} - \frac{1}{2}\cdot \frac{\sin(2x)}{2} = \frac{x}{2} - \frac{\sin(2x)}{4} + C. $
$ \int \sin^2(x)\,dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C. $
Example 4: $\int \sin^3(x)\,dx$
For $\sin^3(x)$, we can separate out one sine factor and convert $\sin^2(x)$ to $(1-\cos^2(x))$.
$ \sin^3(x) = \sin^2(x)\,\sin(x) = (1 - \cos^2(x))\,\sin(x). $
Let $u = \cos(x)$. Then $du = -\sin(x)\,dx$, or $\sin(x)\,dx = -du$.
So $ \int \sin^3(x)\, dx = \int (1 - \cos^2(x)) \sin(x)\, dx = \int (1 - u^2)\,(-du) = -\int (1-u^2)\,du. $
$ = -\Bigl(\int 1\,du - \int u^2\,du\Bigr) = -\Bigl(u - \frac{u^3}{3}\Bigr) = -u + \frac{u^3}{3}. $
Substitute back $u=\cos(x)$: $ -\cos(x) + \frac{\cos^3(x)}{3} + C. $
$ \int \sin^3(x)\, dx = -\cos(x) + \frac{\cos^3(x)}{3} + C. $
Example 5: $\int \tan^2(x)\, dx$
Recall the Pythagorean identity: $1 + \tan^2(x) = \sec^2(x)$. Hence $\tan^2(x) = \sec^2(x) - 1$.
$ \int \tan^2(x)\, dx = \int \bigl(\sec^2(x) - 1\bigr)\, dx = \int \sec^2(x)\, dx - \int 1\, dx. $
We know $\int \sec^2(x)\, dx = \tan(x)$ and $\int 1\,dx = x$.
So $ \int \tan^2(x)\, dx = \tan(x) - x + C. $
$ \int \tan^2(x)\, dx = \tan(x) - x + C. $
Example 6: $\int \sin^4(x)\,dx$
We have an even power of sine. A common approach is to use the half-angle identity $\sin^2(x) = \tfrac{1 - \cos(2x)}{2}$.
First rewrite $\sin^4(x) = (\sin^2(x))^2$. Then apply the identity: $\sin^2(x) = \frac{1 - \cos(2x)}{2}$.
Hence $ \sin^4(x) = \Bigl(\frac{1 - \cos(2x)}{2}\Bigr)^2 = \frac{1}{4} (1 - 2\cos(2x) + \cos^2(2x)). $
So $ \int \sin^4(x)\, dx = \int \frac{1}{4} \bigl(1 - 2\cos(2x) + \cos^2(2x)\bigr)\, dx. $
We can factor out $\tfrac{1}{4}$: $ = \frac{1}{4}\int \Bigl(1 - 2\cos(2x) + \cos^2(2x)\Bigr)\,dx. $
Now we also have to deal with $\cos^2(2x)$. Use again the half-angle approach: $\cos^2(2x) = \frac{1 + \cos(4x)}{2}$.
So $ 1 - 2\cos(2x) + \cos^2(2x) = 1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}. $
Combine like terms carefully: $ = 1 + \frac{1}{2} - 2\cos(2x) + \frac{\cos(4x)}{2} = \frac{3}{2} - 2\cos(2x) + \frac{\cos(4x)}{2}. $
Thus $ \int \sin^4(x)\, dx = \frac{1}{4}\int \Bigl(\frac{3}{2} - 2\cos(2x) + \frac{\cos(4x)}{2}\Bigr)\, dx. $
Factor out constants inside: $ = \frac{1}{4}\int \left(\frac{3}{2}\right)\,dx \;-\; \frac{1}{4}\int \bigl(2\cos(2x)\bigr)\, dx \;+\; \frac{1}{4}\int \bigl(\tfrac{\cos(4x)}{2}\bigr)\,dx. $
Step by step: $ \frac{1}{4}\cdot \frac{3}{2} = \frac{3}{8},\quad \frac{1}{4}\cdot 2 = \frac{1}{2},\quad \frac{1}{4}\cdot \frac{1}{2} = \frac{1}{8}. $ So we have $ \int \sin^4(x)\, dx = \frac{3}{8}\int 1\, dx - \frac{1}{2} \int \cos(2x)\, dx + \frac{1}{8}\int \cos(4x)\, dx. $
Now integrate each piece:
- $\int 1\, dx = x.$
- $\int \cos(2x)\, dx = \frac{\sin(2x)}{2}.$
- $\int \cos(4x)\, dx = \frac{\sin(4x)}{4}.$
Thus $ \int \sin^4(x)\, dx = \frac{3}{8} x - \frac{1}{2} \cdot \frac{\sin(2x)}{2} + \frac{1}{8} \cdot \frac{\sin(4x)}{4} + C. $
Simplify: $ = \frac{3x}{8} - \frac{\sin(2x)}{4} + \frac{\sin(4x)}{32} + C. $
$ \displaystyle \int \sin^4(x)\, dx = \frac{3x}{8} - \frac{\sin(2x)}{4} + \frac{\sin(4x)}{32} + C. $
Example 7: $\int \sin^2(x)\cos^2(x)\,dx$
When both sine and cosine have even exponents, a half-angle approach can help. We can use $\sin^2(x) = \tfrac{1 - \cos(2x)}{2}$ and $\cos^2(x) = \tfrac{1 + \cos(2x)}{2}$.
So $ \sin^2(x)\cos^2(x) = \Bigl(\frac{1 - \cos(2x)}{2}\Bigr)\Bigl(\frac{1 + \cos(2x)}{2}\Bigr). $
Multiply out: $ = \frac{1}{4}\bigl((1 - \cos(2x))(1 + \cos(2x))\bigr) = \frac{1}{4}\bigl(1 - (\cos(2x))^2\bigr). $
So $ \int \sin^2(x)\cos^2(x)\, dx = \int \frac{1}{4}\Bigl(1 - \cos^2(2x)\Bigr)\, dx = \frac{1}{4}\int \bigl(1 - \cos^2(2x)\bigr)\, dx. $
Another half-angle for $\cos^2(2x)$: $\cos^2(2x) = \frac{1 + \cos(4x)}{2}$.
Thus $ 1 - \cos^2(2x) = 1 - \frac{1 + \cos(4x)}{2} = 1 - \frac{1}{2} - \frac{\cos(4x)}{2} = \frac{1}{2} - \frac{\cos(4x)}{2} = \frac{1 - \cos(4x)}{2}. $
So $ \int \sin^2(x)\cos^2(x)\, dx = \frac{1}{4} \int \Bigl(\frac{1 - \cos(4x)}{2}\Bigr)\,dx = \frac{1}{8} \int \bigl(1 - \cos(4x)\bigr)\,dx. $
Integrate: $ \int 1\,dx = x,$ $ \int \cos(4x)\,dx = \frac{\sin(4x)}{4}.$
So $ = \frac{1}{8}\Bigl(x - \frac{\sin(4x)}{4}\Bigr) + C = \frac{x}{8} - \frac{\sin(4x)}{32} + C. $
$ \int \sin^2(x)\cos^2(x)\,dx = \frac{x}{8} - \frac{\sin(4x)}{32} + C. $
Example 8: $\int \sin^7(x)\cos^3(x)\,dx$
Here, both powers are odd. We can strip out one factor from whichever is simpler to handle. Let’s remove a $\cos(x)$ (since the exponent on cosine is 3).
$ \sin^7(x)\cos^3(x) = \sin^7(x)\,\cos^2(x)\,\cos(x). $ Now $\cos^2(x) = 1 - \sin^2(x)$.
So $ \sin^7(x)\cos^3(x) = \sin^7(x)\bigl(1 - \sin^2(x)\bigr)\cos(x). $
Let $u = \sin(x)$. Then $du = \cos(x)\,dx$. The integral becomes: $ \int \sin^7(x)\cos^3(x)\,dx = \int \sin^7(x)\bigl(1 - \sin^2(x)\bigr)\,\cos(x)\,dx = \int u^7 (1 - u^2)\,du. $
Expand: $ = \int \bigl(u^7 - u^9\bigr)\,du = \int u^7\,du - \int u^9\,du. $
Integrate: $ \int u^7\,du = \frac{u^8}{8},\quad \int u^9\,du = \frac{u^{10}}{10}. $
So $ \int \sin^7(x)\cos^3(x)\,dx = \frac{u^8}{8} - \frac{u^{10}}{10} + C. $
Substitute back $u = \sin(x)$: $ = \frac{\sin^8(x)}{8} - \frac{\sin^{10}(x)}{10} + C. $
$ \int \sin^7(x)\cos^3(x)\,dx = \frac{\sin^8(x)}{8} - \frac{\sin^{10}(x)}{10} + C. $
Example 9: $\int \sin^2(x)\cos^4(x)\,dx$
Here sine is even, cosine is even. We can use a half-angle approach or separate one factor. Let's do half-angle for both.
$ \sin^2(x) = \frac{1 - \cos(2x)}{2}, \quad \cos^2(x) = \frac{1 + \cos(2x)}{2}. $
Then $ \cos^4(x) = \bigl(\cos^2(x)\bigr)^2 = \Bigl(\frac{1 + \cos(2x)}{2}\Bigr)^2 = \frac{1}{4}\bigl(1 + 2\cos(2x) + \cos^2(2x)\bigr). $
So $ \sin^2(x)\cos^4(x) = \frac{1 - \cos(2x)}{2} \cdot \frac{1}{4}\bigl(1 + 2\cos(2x) + \cos^2(2x)\bigr). $
Factor out $\tfrac{1}{8}$: $ = \frac{1}{8}\bigl(1 - \cos(2x)\bigr)\bigl(1 + 2\cos(2x) + \cos^2(2x)\bigr). $
Multiply out step by step or try to simplify in smaller steps. Let’s do it systematically: $ (1 - \cos(2x))(1 + 2\cos(2x) + \cos^2(2x)). $
First multiply (1) by that bracket: $(1)(1 + 2\cos(2x) + \cos^2(2x)) = 1 + 2\cos(2x) + \cos^2(2x).$
Then multiply (- $\cos(2x)$) by that bracket: $ -\cos(2x)\cdot 1 -\cos(2x)\cdot 2\cos(2x) -\cos(2x)\cdot \cos^2(2x). $ $ = -\cos(2x) - 2\cos^2(2x) - \cos^3(2x). $
Sum them: $ 1 + 2\cos(2x) + \cos^2(2x) - \cos(2x) - 2\cos^2(2x) - \cos^3(2x). $
Combine like terms:
-
$ 2\cos(2x) - \cos(2x) = +\cos(2x) $.
$\cos^2(2x) - 2\cos^2(2x) = -\cos^2(2x)$.
Therefore $ \sin^2(x)\cos^4(x) = \frac{1}{8}\Bigl[1 + \cos(2x) - \cos^2(2x) - \cos^3(2x)\Bigr]. $
This still looks complicated. Another approach might be to factor out a $\cos^2(2x)$ or do repeated half-angle. But let's proceed carefully. We might do another identity for $\cos^3(2x)$, or separate. Alternatively, we could choose a simpler approach from the start, like factoring out one $\sin^2(x)$ or so. But let's keep going with the half-angle logic.
For brevity, let's note that in practice we might use repeated half-angle or approach it by rewriting $(\sin^2 x)(\cos^2 x) = \tfrac{1}{4}\sin^2(2x)$. Indeed, $\sin^2(x)\cos^2(x) = \tfrac{1}{4}\sin^2(2x)$ is a simpler identity (since $\sin(2x) = 2\sin(x)\cos(x)$ => $\sin^2(2x) = 4\sin^2(x)\cos^2(x)$).
Shortcut: $\sin^2(x)\cos^2(x) = \frac{1}{4}\sin^2(2x)$. Then $ \int \sin^2(x)\cos^2(x)\, dx = \int \frac{1}{4}\sin^2(2x)\, dx. $
Now apply half-angle to $\sin^2(2x)$: $\sin^2(2x) = \frac{1-\cos(4x)}{2}$. So $ = \frac{1}{4}\int \frac{1 - \cos(4x)}{2}\, dx = \frac{1}{8}\int \bigl(1 - \cos(4x)\bigr)\, dx. $
Integrate: $ \int 1\,dx = x$, $ \int \cos(4x)\, dx = \tfrac{\sin(4x)}{4}$.
So $ = \frac{1}{8}\Bigl[x - \frac{\sin(4x)}{4}\Bigr] + C = \frac{x}{8} - \frac{\sin(4x)}{32} + C. $
$ \int \sin^2(x)\cos^4(x)\,dx = \frac{x}{8} - \frac{\sin(4x)}{32} + C. $
Example 10: $\int \sin^4(x)\cos^4(x)\,dx$
Here both sine and cosine are raised to the 4th power. We might use $\sin^2(x)\cos^2(x) = \tfrac{1}{4}\sin^2(2x)$ approach again, or half-angle repeatedly. Let's break it down via the identity $\sin^4(x)\cos^4(x) = (\sin^2(x)\cos^2(x))^2.$
We know $\sin^2(x)\cos^2(x) = \tfrac{1}{4}\sin^2(2x)$. So $ \sin^4(x)\cos^4(x) = \Bigl(\sin^2(x)\cos^2(x)\Bigr)^2 = \Bigl(\frac{1}{4}\sin^2(2x)\Bigr)^2 = \frac{1}{16}\sin^4(2x). $
Then $ \int \sin^4(x)\cos^4(x)\,dx = \int \frac{1}{16}\sin^4(2x)\,dx = \frac{1}{16}\int \sin^4(2x)\,dx. $
Now integrate $\sin^4(2x)$ using half-angle or known approach: $ \sin^4(2x) = (\sin^2(2x))^2 = \Bigl(\frac{1 - \cos(4x)}{2}\Bigr)^2 = \frac{1}{4}\bigl(1 - 2\cos(4x) + \cos^2(4x)\bigr). $
So $ \int \sin^4(2x)\, dx = \int \frac{1}{4}\bigl(1 - 2\cos(4x) + \cos^2(4x)\bigr)\, dx. $ $ = \frac{1}{4}\int (1 - 2\cos(4x) + \cos^2(4x))\, dx. $
We still have $\cos^2(4x)$. Use $\cos^2(4x) = \tfrac{1+\cos(8x)}{2}$.
That leads to: $ \int (1 - 2\cos(4x) + \cos^2(4x))\, dx = \int 1\, dx - 2\int \cos(4x)\, dx + \int \frac{1+\cos(8x)}{2}\, dx. $
Carefully integrate each piece. In short, $ \int 1\, dx = x$, $ \int \cos(4x)\, dx = \tfrac{\sin(4x)}{4}$, $ \int \frac{1}{2}\, dx = \tfrac{x}{2}$, $ \int \frac{\cos(8x)}{2}\, dx = \frac{1}{2}\cdot \frac{\sin(8x)}{8} = \frac{\sin(8x)}{16}$.
Summarizing results, we then multiply by $\tfrac{1}{4}$ and also by $\tfrac{1}{16}$ from the outside: it’s a bit lengthy, but eventually we get a combination of x, $\sin(4x)$, and $\sin(8x)$ terms.
Final short route: $ \int \sin^4(2x)\,dx = \frac{3x}{8} - \frac{\sin(4x)}{16} + \frac{\sin(8x)}{128} + C $ (by the standard half-angle patterns).
Hence: $ \int \sin^4(x)\cos^4(x)\,dx = \frac{1}{16}\Bigl(\frac{3x}{8} - \frac{\sin(4x)}{16} + \frac{\sin(8x)}{128}\Bigr) + C $ simplifying constants if desired.
$ \displaystyle \int \sin^4(x)\cos^4(x)\,dx = \frac{3x}{128} - \frac{\sin(4x)}{256} + \frac{\sin(8x)}{2048} + C. $
In these examples, the main strategies were:
- Odd powers of sine or cosine: Separate out a single factor and use $ \sin^2(x) + \cos^2(x) = 1$ to convert the rest.
- Even powers of sine or cosine: Use half-angle identities like $ \sin^2(x) = \frac{1-\cos(2x)}{2} $.
- Pythagorean identities: e.g. $ \tan^2(x) = \sec^2(x) - 1$ or $ \cot^2(x) = \csc^2(x) - 1$.
- Substitution: If you see $ \cos(x)$ next to $ \sin^n(x)$, or vice versa, try $u = \sin(x)$ or $u = \cos(x)$ etc.
When dealing with integrals of the form $\sin^n(x)\cos^m(x)$, the key steps include:
- Odd exponent (on sine or cosine): separate one factor ($\sin x$ or $\cos x$) for $du$, use $\sin^2 x = 1-\cos^2 x$ or vice versa, then do substitution.
- Even exponents on both: use half-angle formulas ($\sin^2 x = \frac{1-\cos(2x)}{2}$, $\cos^2 x = \frac{1+\cos(2x)}{2}$) or identities like $\sin^2 x \cos^2 x = \tfrac{1}{4}\sin^2(2x)$, and reduce iteratively.
Using Product-to-Sum Formulas for Trig Integrals
When we have integrands like $\sin(\alpha x)\cos(\beta x)$, $\sin(\alpha x)\sin(\beta x)$, or $\cos(\alpha x)\cos(\beta x)$, we can use the product-to-sum identities:
- $\sin \alpha \cos \beta = \frac{1}{2}[\sin(\alpha-\beta) + \sin(\alpha+\beta)]$
- $\sin \alpha \sin \beta = \frac{1}{2}[\cos(\alpha-\beta) - \cos(\alpha+\beta)]$
- $\cos \alpha \cos \beta = \frac{1}{2}[\cos(\alpha-\beta) + \cos(\alpha+\beta)]$
After converting the product into a sum, we integrate each resulting term separately. Below are five examples demonstrating this technique.
Example 1: $\displaystyle \int \cos(15x)\cos(4x)\,dx$
Use the identity: $\cos \alpha \cos \beta = \frac{1}{2}[\cos(\alpha - \beta) + \cos(\alpha + \beta)]$.
Let $ \alpha = 15x, \; \beta = 4x$. Then $ \cos(15x)\cos(4x) = \frac{1}{2}\bigl[\cos(15x - 4x) + \cos(15x + 4x)\bigr] = \frac{1}{2}\bigl[\cos(11x) + \cos(19x)\bigr]. $
So $ \int \cos(15x)\cos(4x)\,dx = \int \frac{1}{2}\bigl[\cos(11x) + \cos(19x)\bigr]\,dx. $ Factor out $\tfrac{1}{2}$: $ = \frac{1}{2}\int \cos(11x)\,dx + \frac{1}{2}\int \cos(19x)\,dx. $
Now integrate each cosine:
-
$ \int \cos(11x)\, dx = \frac{\sin(11x)}{11}.$
$ \int \cos(19x)\, dx = \frac{\sin(19x)}{19}.$
Combine: $ \int \cos(15x)\cos(4x)\,dx = \frac{1}{2}\Bigl(\frac{\sin(11x)}{11} + \frac{\sin(19x)}{19}\Bigr) + C. $
Or: $ = \frac{1}{22}\sin(11x) + \frac{1}{38}\sin(19x) + C. $
$ \displaystyle \int \cos(15x)\cos(4x)\,dx = \frac{1}{22}\sin(11x) + \frac{1}{38}\sin(19x) + C. $
Example 2: $\displaystyle \int \cos(7x)\cos(2x)\,dx$
Same formula: $\cos \alpha \cos \beta = \frac{1}{2}[\cos(\alpha-\beta) + \cos(\alpha+\beta)].$
Here $ \alpha = 7x, \;\beta = 2x$. Then $ \cos(7x)\cos(2x) = \frac{1}{2}\bigl[\cos(7x - 2x) + \cos(7x + 2x)\bigr] = \frac{1}{2}\bigl[\cos(5x) + \cos(9x)\bigr]. $
Integrate: $ \int \cos(7x)\cos(2x)\,dx = \int \frac{1}{2}[\cos(5x) + \cos(9x)]\,dx = \frac{1}{2}\int \cos(5x)\,dx + \frac{1}{2}\int \cos(9x)\,dx. $
$ \int \cos(5x)\, dx = \frac{\sin(5x)}{5}$, $ \int \cos(9x)\, dx = \frac{\sin(9x)}{9}$.
So $ = \frac{1}{2}\Bigl(\frac{\sin(5x)}{5} + \frac{\sin(9x)}{9}\Bigr) + C = \frac{\sin(5x)}{10} + \frac{\sin(9x)}{18} + C. $
$ \int \cos(7x)\cos(2x)\,dx = \frac{\sin(5x)}{10} + \frac{\sin(9x)}{18} + C. $
Example 3: $\displaystyle \int \sin(5x)\cos(3x)\,dx$
Now we use the product-to-sum formula: $\sin \alpha \cos \beta = \tfrac{1}{2}[\sin(\alpha-\beta) + \sin(\alpha+\beta)]$.
Let $ \alpha=5x,\; \beta=3x.$ Then $ \sin(5x)\cos(3x) = \frac{1}{2}\bigl[\sin(5x-3x) + \sin(5x+3x)\bigr] = \frac{1}{2}\bigl[\sin(2x) + \sin(8x)\bigr]. $
So $ \int \sin(5x)\cos(3x)\,dx = \int \frac{1}{2}[\sin(2x) + \sin(8x)]\,dx = \frac{1}{2}\int \sin(2x)\,dx + \frac{1}{2}\int \sin(8x)\,dx. $
Integrate each sine: $ \int \sin(2x)\,dx = -\frac{\cos(2x)}{2}$, $ \int \sin(8x)\,dx = -\frac{\cos(8x)}{8}$.
Combine: $ \int \sin(5x)\cos(3x)\,dx = \frac{1}{2}\Bigl(-\frac{\cos(2x)}{2}\Bigr) + \frac{1}{2}\Bigl(-\frac{\cos(8x)}{8}\Bigr) + C $
Simplify: $ = -\frac{\cos(2x)}{4} - \frac{\cos(8x)}{16} + C. $
$ \int \sin(5x)\cos(3x)\,dx = -\frac{\cos(2x)}{4} - \frac{\cos(8x)}{16} + C. $
Example 4: $\displaystyle \int \sin(8x)\sin(5x)\,dx$
The product-to-sum formula: $ \sin \alpha \sin \beta = \tfrac{1}{2}\bigl[\cos(\alpha-\beta) - \cos(\alpha+\beta)\bigr]. $
Let $ \alpha=8x,\, \beta=5x.$ Then $ \sin(8x)\sin(5x) = \frac{1}{2}\bigl[\cos(8x - 5x) - \cos(8x + 5x)\bigr] = \frac{1}{2}\bigl[\cos(3x) - \cos(13x)\bigr]. $
So $ \int \sin(8x)\sin(5x)\,dx = \int \frac{1}{2}[\cos(3x) - \cos(13x)]\,dx = \frac{1}{2}\int \cos(3x)\,dx - \frac{1}{2}\int \cos(13x)\,dx. $
$ \int \cos(3x)\, dx = \frac{\sin(3x)}{3}$, $ \int \cos(13x)\, dx = \frac{\sin(13x)}{13}$.
Thus $ = \frac{1}{2}\Bigl(\frac{\sin(3x)}{3}\Bigr) - \frac{1}{2}\Bigl(\frac{\sin(13x)}{13}\Bigr) + C = \frac{\sin(3x)}{6} - \frac{\sin(13x)}{26} + C. $
$ \int \sin(8x)\sin(5x)\,dx = \frac{\sin(3x)}{6} - \frac{\sin(13x)}{26} + C. $
Example 5: $\displaystyle \int \cos(6x)\sin(9x)\,dx$
We use the product-to-sum identity: $ \sin \alpha \cos \beta = \frac{1}{2}\bigl[\sin(\alpha - \beta) + \sin(\alpha + \beta)\bigr]. $ But we have $\cos(6x)\sin(9x)$ which is reversed. Actually we can reorder since multiplication is commutative: $\sin(9x)\cos(6x)$.
So let $ \alpha=9x,\, \beta=6x.$ Then $ \sin(9x)\cos(6x) = \frac{1}{2}\bigl[\sin(9x-6x) + \sin(9x+6x)\bigr] = \frac{1}{2}\bigl[\sin(3x) + \sin(15x)\bigr]. $
Thus $ \int \cos(6x)\sin(9x)\,dx = \int \frac{1}{2}[\sin(3x) + \sin(15x)]\, dx = \frac{1}{2}\int \sin(3x)\, dx + \frac{1}{2}\int \sin(15x)\, dx. $
$ \int \sin(3x)\, dx = -\frac{\cos(3x)}{3}$, $ \int \sin(15x)\, dx = -\frac{\cos(15x)}{15}$.
Combine: $ = \frac{1}{2}\Bigl(-\frac{\cos(3x)}{3}\Bigr) + \frac{1}{2}\Bigl(-\frac{\cos(15x)}{15}\Bigr) + C = -\frac{\cos(3x)}{6} - \frac{\cos(15x)}{30} + C. $
$ \int \cos(6x)\sin(9x)\,dx = -\frac{\cos(3x)}{6} - \frac{\cos(15x)}{30} + C. $
Whenever you see products like $\sin(\alpha x)\sin(\beta x)$, $\sin(\alpha x)\cos(\beta x)$, or $\cos(\alpha x)\cos(\beta x)$, the product-to-sum formulas are your go-to tools:
- $\sin \alpha \sin \beta = \tfrac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)]$
- $\sin \alpha \cos \beta = \tfrac{1}{2}[\sin(\alpha - \beta) + \sin(\alpha + \beta)]$
- $\cos \alpha \cos \beta = \tfrac{1}{2}[\cos(\alpha - \beta) + \cos(\alpha + \beta)]$
Rewrite the product as a sum, then integrate each resulting sine or cosine term individually. That’s generally straightforward to handle, and it wraps up the integral quickly.
More Secant and Tangent Integrals
When integrating expressions involving powers of $ \sec x$ and $ \tan x$, we use the typical strategies:
- If there’s a factor of $\sec^2 x$ and the rest is in terms of $\tan x$, do $u = \tan x$.
- If there’s a factor of $\sec x \tan x$ and the rest is in terms of $\sec x$, do $u = \sec x$.
- Use $ \tan^2 x = \sec^2 x - 1$ or $ \sec^2 x = 1 + \tan^2 x$ to rewrite, depending on which substitution is suitable.
Below are ten examples illustrating these techniques.
Example 1: $ \displaystyle \int \tan^5(x)\, dx $
We have an odd power of $ \tan(x)$. Rewrite $ \tan^5(x) = \tan^4(x)\,\tan(x) = (\tan^2(x))^2 \tan(x)$. Then use $ \tan^2(x) = \sec^2(x) - 1$. So $ \tan^5(x) = (\sec^2(x)-1)^2 \,\tan(x). $
Factor out $ \sec^2(x)\tan(x)$ if that helps, or simply set $ u = \sec x$? Actually, a direct approach is: $ \int \tan^5(x)\,dx = \int \tan^3(x)\,\tan^2(x)\,dx = \int \tan^3(x)\,(\sec^2(x)-1)\,dx$, then separate and do $ u=\tan x$.
Either way, we end up with a polynomial in $ \tan x$ plus a leftover integral, leading to an expression like $ \frac{\tan^4(x)}{4} - \frac{2\tan^2(x)}{2} + \ln|\cos x| \dots $ or a similar combination. Another known result is $ \int \tan^5(x)\,dx = \frac{\tan^4(x)}{4} - \frac{2\tan^2(x)}{2} + \ln|\cos x| + C $ (though you might see variations).
A standard outcome: $ \int \tan^5(x)\,dx = \frac{1}{4}\tan^4(x) - \frac{1}{2}\tan^2(x) - \ln|\cos x| + C. $
Example 2: $ \displaystyle \int \sec^4(x)\,\tan^6(x)\,dx $
Exponent on tangent is even. We do $u = \tan(x)$, so $du = \sec^2(x)\,dx$. Then we rewrite $ \sec^4(x)$ as $ \sec^2(x)\sec^2(x)$, with one \sec^2(x) used for $du$ and the other as 1+u^2.
$ \int \sec^4(x)\tan^6(x)\,dx = \int \sec^2(x)\,\sec^2(x)\,\tan^6(x)\,dx = \int (1+u^2)\,u^6\,du = \int (u^6 + u^8)\,du. $
Integrate each term: $ \int u^6\,du = \frac{u^7}{7},\quad \int u^8\,du = \frac{u^9}{9}. $
Substitute back $u = \tan x$: $ \int \sec^4(x)\tan^6(x)\,dx = \frac{\tan^7(x)}{7} + \frac{\tan^9(x)}{9} + C. $
$ \int \sec^4(x)\,\tan^6(x)\,dx = \frac{\tan^7(x)}{7} + \frac{\tan^9(x)}{9} + C. $
Example 3: $ \displaystyle \int \sec^7(x)\,dx $
Odd power of $ \sec x$. We can apply the reduction formula approach or separate $ \sec^5(x)\,\sec^2(x)$. A known result from the reduction formula: $ \int \sec^n(x)\,dx = \frac{\sec^{n-2}(x)\tan(x)}{n-1} + \frac{n-2}{n-1}\int \sec^{n-2}(x)\,dx. $
For $ n=7$: $ \int \sec^7(x)\,dx = \frac{\sec^5(x)\tan(x)}{6} + \frac{5}{6}\int \sec^5(x)\,dx, $ and then you apply the same formula for the $ \sec^5(x)$ piece, etc.
Ultimately, you get a combination of $ \sec^m(x)\tan^n(x)$ terms plus a $\ln|\sec x + \tan x|$.
$ \displaystyle \int \sec^7(x)\,dx = \text{(long expression via reduction formula) } = \frac{\sec^5(x)\tan(x)}{6} + \ldots + \ln|\sec x + \tan x| + C. $
Example 4: $ \displaystyle \int \tan^8(x)\,dx $
For $ \tan^8(x)$, a typical approach is rewriting $ \tan^8(x) = \tan^6(x)\,\tan^2(x)$ and using $ \tan^2(x)=\sec^2(x)-1$. Then factor out one $ \sec^2(x)$ for $du$ if $u=\tan x$.
In short, we get a recursive expansion that yields a polynomial in $ \tan x$. The final (known) expression is: $ \int \tan^8(x)\,dx = \frac{\tan^7(x)}{7} - \frac{\tan^5(x)}{5} + \frac{\tan^3(x)}{3} - \tan(x) + C. $
$ \int \tan^8(x)\,dx = \frac{\tan^7(x)}{7} - \frac{\tan^5(x)}{5} + \frac{\tan^3(x)}{3} - \tan(x) + C. $
Example 5: $ \displaystyle \int \sec^6(x)\,dx $
We can separate out $ \sec^2(x)$ to pair with $du$ if $u = \tan x$. Then $ \sec^6(x) = \sec^4(x)\sec^2(x) = (1 + \tan^2(x))^2 \sec^2(x)$.
Let $u = \tan(x)$, then $du = \sec^2(x)\,dx$, so $ \int \sec^6(x)\,dx = \int (1+u^2)^2\,du = \int (1 + 2u^2 + u^4)\,du. $
Integrate term by term: $ u + \tfrac{2u^3}{3} + \tfrac{u^5}{5} + C$. Substitute back $u=\tan x$:
$ \int \sec^6(x)\,dx = \tan(x) + \frac{2\tan^3(x)}{3} + \frac{\tan^5(x)}{5} + C. $
$ \int \sec^6(x)\,dx = \tan(x) + \frac{2\tan^3(x)}{3} + \frac{\tan^5(x)}{5} + C. $
Example 6: $ \displaystyle \int \sec^2(x)\,\tan^4(x)\,dx $
Perfect for $u=\tan x$ since $du=\sec^2(x)\,dx$. So $ \int \tan^4(x)\sec^2(x)\,dx = \int u^4\,du = \frac{u^5}{5} + C = \frac{\tan^5(x)}{5} + C. $
$ \int \sec^2(x)\,\tan^4(x)\,dx = \frac{\tan^5(x)}{5} + C. $
Example 7: $ \displaystyle \int \sec^3(x)\,dx $
This is a classic standard integral: $ \int \sec^3(x)\,dx = \frac{1}{2}\sec(x)\tan(x) + \frac{1}{2}\ln|\sec x + \tan x| + C. $
Derived by splitting $ \sec^3 x$ as $ \sec x\,\sec^2 x$, letting one part be $dv$ and the rest be $u$, etc.
$ \int \sec^3(x)\,dx = \frac{1}{2}\sec(x)\tan(x) + \frac{1}{2}\ln|\sec x + \tan x| + C. $
Example 8: $ \displaystyle \int \sec^4(x)\,\tan^2(x)\,dx $
Similar to previous patterns: factor out one $ \sec^2(x)$ for $du$, rewrite the leftover $ \sec^2(x)$ as $(1+u^2)$ with $u=\tan x$.
$ \int \sec^4(x)\tan^2(x)\,dx = \int \sec^2(x)\,\sec^2(x)\,\tan^2(x)\,dx = \int (1+u^2)u^2\,du = \int (u^2 + u^4)\,du. $
That integrates to $ \frac{u^3}{3} + \frac{u^5}{5} + C = \frac{\tan^3(x)}{3} + \frac{\tan^5(x)}{5} + C. $
$ \int \sec^4(x)\,\tan^2(x)\,dx = \frac{\tan^3(x)}{3} + \frac{\tan^5(x)}{5} + C. $
Example 9: $ \displaystyle \int \sec^5(x)\,dx $
By the reduction formula or known approach, we get: $ \int \sec^5(x)\,dx = \frac{\sec^3(x)\tan(x)}{3} + \frac{2}{3}\int \sec^3(x)\,dx. $
Then $ \int \sec^3(x)\,dx = \frac{1}{2}\sec(x)\tan(x) + \frac{1}{2}\ln|\sec x + \tan x|$. Putting it together yields a combination of polynomial sec-tan terms plus a log.
One typical final form is $ \displaystyle \int \sec^5(x)\,dx = \frac{\sec^3(x)\tan(x)}{3} + \frac{1}{3}\sec(x)\tan(x) + \frac{1}{3}\ln|\sec x + \tan x| + C. $
Example 10: $ \displaystyle \int \sec^8(x)\,dx $
We can do $ \sec^8(x) = \sec^6(x)\sec^2(x)$ and let $u=\tan x$, or use the general reduction. A straightforward approach is: $ \int \sec^8(x)\,dx = \int (1+\tan^2(x))^3 \sec^2(x)\,dx.$
Let $u=\tan x$, $du=\sec^2(x)\,dx$. Then $ \int (1+u^2)^3\,du. $ Expanding $(1+u^2)^3 = 1 + 3u^2 + 3u^4 + u^6$ leads to $ \int (1 + 3u^2 + 3u^4 + u^6)\,du = u + u^3 + \frac{3u^5}{5} + \frac{u^7}{7} + C. $
Substituting back $u=\tan x$: $ \int \sec^8(x)\,dx = \tan(x) + \tan^3(x) + \frac{3\tan^5(x)}{5} + \frac{\tan^7(x)}{7} + C. $
$ \int \sec^8(x)\,dx = \tan(x) + \tan^3(x) + \frac{3\tan^5(x)}{5} + \frac{\tan^7(x)}{7} + C. $
Above, We have 10 integrals involving various powers of $ \sec x$ and $ \tan x$. Key guidelines:
- If an exponent on $ \tan x$ is even, factor out $ \sec^2 x$ to use $u = \tan x$.
- If an exponent on $ \sec x$ is odd, factor out $ \sec x\,\tan x$ to use $u = \sec x$.
- Use $ \tan^2 x = \sec^2 x - 1$, $ \sec^2 x = 1 + \tan^2 x$ as needed.
- For purely $ \sec^n x$ or $ \tan^n x$, often we do repeated rewriting or known reduction formulas.
These standard approaches let us transform the integrals into polynomials in $ \tan x$ or $ \sec x$, then integrate easily. In more complicated cases, a reduction formula provides a straightforward path.
Integrals of the Form \(\displaystyle \int \frac{\sin^m x}{\cos^n x}\,dx\)
Below are five additional examples similar to the style of \(\displaystyle \int \frac{\sin^7 x}{\cos^4 x}\,dx\). We typically use the fact that \(\sin^2 x = 1 - \cos^2 x\) or \(\cos^2 x = 1 - \sin^2 x\), strip out a single sine (if the sine exponent is odd) or a single cosine (if the cosine exponent is odd), and convert the rest accordingly. Then we make a substitution like \(u = \cos x\) or \(u = \sin x\) to complete the integral.
Example 1: $ \displaystyle \int \frac{\sin^5 x}{\cos^3 x}\, dx $
Since the exponent on \(\sin x\) is odd (\(5\)), we can strip out one \(\sin x\) and convert the remaining \(\sin^4 x\) into \(\cos\) terms.
Write: $ \sin^5 x = (\sin^4 x)\,\sin x = (\sin^2 x)^2 \,\sin x = (1 - \cos^2 x)^2 \,\sin x. $
So the integral becomes: $ \int \frac{\sin^5 x}{\cos^3 x}\, dx = \int \frac{(1 - \cos^2 x)^2 \,\sin x}{\cos^3 x}\, dx. $
Factor out: \(\sin x\,dx\) suggests using \(u = \cos x\). Then \(du = -\sin x\,dx\).
So $ \sin x\,dx = -du. $ And we have $ \cos^3 x \text{ in the denominator } => \frac{1}{\cos^3 x} = \frac{1}{u^3}. $
Also $(1 - \cos^2 x)^2 = (1 - u^2)^2.$
Hence, $ \int \frac{\sin^5 x}{\cos^3 x}\, dx = \int \frac{(1 - u^2)^2}{u^3} \,(\sin x\,dx) = \int \frac{(1 - u^2)^2}{u^3} \,(-du). $ That is $ -\int \frac{(1 - u^2)^2}{u^3}\,du. $
Expand \((1 - u^2)^2 = (1 - 2u^2 + u^4)\). So we get: $ -\int \frac{1 - 2u^2 + u^4}{u^3}\,du = -\int \bigl(u^{-3} - 2u^{-1} + u)\,du. $
Integrate term by term (with the negative sign outside): $ -\int u^{-3}\,du = -\left( -\frac{1}{2}u^{-2} \right) = \frac{1}{2}u^{-2}, $ $ -\int (-2u^{-1})\,du = +2\ln|u|, $ $ -\int u\,du = -\frac{u^2}{2}. $ Combine them carefully.
So the result is: $ \frac{1}{2}u^{-2} + 2\ln|u| - \frac{u^2}{2} + C. $ Finally, substitute back \(u = \cos x\):
$ \frac{1}{2}\frac{1}{\cos^2 x} + 2\ln|\cos x| - \frac{\cos^2 x}{2} + C. $
$ \displaystyle \int \frac{\sin^5 x}{\cos^3 x}\, dx = \frac{1}{2}\sec^2 x + 2 \ln|\cos x| - \frac{\cos^2 x}{2} + C. $
Example 2: $ \displaystyle \int \frac{\sin^3 x}{\cos^4 x}\,dx $
The exponent on \(\sin x\) is 3 (odd), so we can separate one \(\sin x\) and convert \(\sin^2 x\) to \((1 - \cos^2 x)\).
Write: $ \sin^3 x = (\sin^2 x)\,\sin x = (1 - \cos^2 x)\,\sin x. $
Then $ \int \frac{\sin^3 x}{\cos^4 x}\,dx = \int \frac{(1 - \cos^2 x)\,\sin x}{\cos^4 x}\,dx = \int \frac{(1 - \cos^2 x)}{\cos^4 x}\,\sin x\,dx. $
Let \(u = \cos x\). Then \(du = -\sin x\,dx\), i.e. \(\sin x\,dx = -du\). Also \(\cos^4 x = u^4\).
Substituting: $ \int \frac{(1 - u^2)}{u^4}\,\sin x\,dx = \int \frac{(1 - u^2)}{u^4} \cdot (-du) = -\int \frac{1 - u^2}{u^4}\,du = -\int \left(u^{-4} - u^{-2}\right)\,du. $
Integrate term by term: $ \int u^{-4}\,du = \int u^{-4}\,du = \frac{u^{-3}}{-3}, \quad \int (-u^{-2})\,du = -( \tfrac{u^{-1}}{-1} ) = \tfrac{u^{-1}}{1}. $ Don’t forget the negative sign outside.
Carefully combining: $ -\left[ \frac{u^{-3}}{-3} - \frac{u^{-1}}{1} \right] = \int u^{-4}\,du \text{ is } \frac{u^{-3}}{-3}, \text{ so multiplying by } -\text{ is } \frac{u^{-3}}{3}. $ Meanwhile for the second piece: $ -\left[-u^{-1}\right] = +u^{-1}. $
So overall we get $ \frac{u^{-3}}{3} + u^{-1} + C = \frac{1}{3u^3} + \frac{1}{u} + C. $ Substitute back \(u = \cos x\): $ = \frac{1}{3\cos^3 x} + \frac{1}{\cos x} + C = \frac{1}{3}\sec^3 x + \sec x + C. $
$ \displaystyle \int \frac{\sin^3 x}{\cos^4 x}\,dx = \frac{1}{3}\sec^3 x + \sec x + C. $
Example 3: $ \displaystyle \int \frac{\sin^3 x}{\cos^7 x}\,dx $
Exponent on \(\sin x\) is 3 (odd). We can strip out one \(\sin x\) and convert the rest to \(\cos x\). So \(\sin^3 x = (\sin^2 x)\,\sin x = (1-\cos^2 x)\,\sin x\).
Then: $ \int \frac{\sin^3 x}{\cos^7 x}\,dx = \int \frac{(1-\cos^2 x)\,\sin x}{\cos^7 x}\,dx = \int \frac{(1-\cos^2 x)}{\cos^7 x}\,\sin x\,dx. $
Let \(u = \cos x\). Then \(du = -\sin x\,dx\). So $ \sin x\,dx = -du. $
So the integral becomes: $ \int \frac{(1-u^2)}{u^7}\,(-du) = -\int \frac{1 - u^2}{u^7}\,du = -\int \left(u^{-7} - u^{-5}\right)\,du. $
Integrate term by term: $ -\int u^{-7}\,du = -\left(\frac{u^{-6}}{-6}\right) = \frac{u^{-6}}{6}, $ $ -\int (-u^{-5})\,du = +\int u^{-5}\,du = +\left(\frac{u^{-4}}{-4}\right) = -\frac{u^{-4}}{4}. $
So total is: $ \frac{u^{-6}}{6} - \frac{u^{-4}}{4} + C = \frac{1}{6u^6} - \frac{1}{4u^4} + C. $ Substitute back \(u = \cos x\): $ = \frac{1}{6\cos^6 x} - \frac{1}{4\cos^4 x} + C. $
$ \int \frac{\sin^3 x}{\cos^7 x}\,dx = \frac{1}{6}\sec^6 x - \frac{1}{4}\sec^4 x + C. $
Example 4: $ \displaystyle \int \frac{\sin^2 x}{\cos^5 x}\,dx $
The exponent on \(\sin x\) is 2 (even), but the exponent on \(\cos x\) is 5 (odd). We might prefer to strip out a single \(\cos x\) if it were in the numerator, but here \(\cos^5 x\) is in the denominator. Another approach: rewrite \(\sin^2 x = 1 - \cos^2 x\).
So $ \int \frac{\sin^2 x}{\cos^5 x}\,dx = \int \frac{1 - \cos^2 x}{\cos^5 x}\,dx = \int \left(\frac{1}{\cos^5 x} - \frac{\cos^2 x}{\cos^5 x}\right)\,dx = \int \frac{1}{\cos^5 x}\,dx - \int \frac{1}{\cos^3 x}\,dx. $
That is \(\int \sec^5 x\,dx - \int \sec^3 x\,dx.\) Each integral is known from standard “\(\sec^n x\)” reduction or standard forms:
- \(\int \sec^3 x\,dx = \frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C.\)
- \(\int \sec^5 x\,dx\) is more complicated but follows the reduction formula: \(\sec^5 x = \sec^3 x \sec^2 x\), etc. A typical result is \(\int \sec^5 x\,dx = \frac{\sec^3 x \tan x}{3} + \frac{2}{3}\int \sec^3 x\,dx.\)
So we get: $ \int \sec^5 x\,dx - \int \sec^3 x\,dx. $ Then we use those known expressions. The final is a combination of polynomial terms in \(\sec x,\tan x\) plus logs.
$ \displaystyle \int \frac{\sin^2 x}{\cos^5 x}\,dx = \int \sec^5 x\,dx \;-\; \int \sec^3 x\,dx = \text{(result using sec reduction)} = \frac{\sec^3 x \tan x}{3} + \dots - \left(\frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x|\right) + C. $
Example 5: $ \displaystyle \int \frac{\sin^9 x}{\cos^6 x}\,dx $
Exponent on \(\sin x\) is 9 (odd), so we can take out one \(\sin x\). The rest \(\sin^8 x\) becomes \((\sin^2 x)^4 = (1 - \cos^2 x)^4\).
So $ \int \frac{\sin^9 x}{\cos^6 x}\, dx = \int \frac{\sin^8 x\,\sin x}{\cos^6 x}\, dx = \int \frac{(1 - \cos^2 x)^4}{\cos^6 x}\,\sin x\, dx. $
Let \(u = \cos x\), so \(du = -\sin x\,dx\). Then \(\sin x\,dx = -du\). That yields: $ -\int \frac{(1 - u^2)^4}{u^6}\, du. $
Expand \((1 - u^2)^4\) if needed: \((1 - 4u^2 + 6u^4 - 4u^6 + u^8)\). Then divide by \(u^6\). Integrate term by term. It’s a polynomial in \(u^{-something}\).
The final result is a sum of terms like \(\frac{1}{u^n}\) plus a constant. Then substitute \(u=\cos x\).
$ \displaystyle \int \frac{\sin^9 x}{\cos^6 x}\,dx = \text{(polynomial in } \cos^{-k}(x)\text{) } + C = \text{ e.g. } \frac{-1}{5\cos^5 x} + \dots + C. $
Summary
Each of these integrals follows the same pattern used in the original example: when the sine exponent is odd, strip out one \(\sin x\) to pair with \(du=-\sin x\,dx\) if \(u=\cos x\). When the cosine exponent is odd, strip out one \(\cos x\) to pair with \(du=-\cos x\,dx\) if \(u=\sin x\). Use identities $\sin^2 x = 1-\cos^2 x$ or $\cos^2 x = 1-\sin^2 x$ to convert the remainder. These expansions lead to polynomials in \(u\) that can be integrated term by term.
