## 17Calculus - Sine-Cosine Trig Integration

This page covers integration of functions involving sines and/or cosines in more advanced form that require techniques other than just integration by substitution. [If you are first learning sine and cosine in integration, check out the basics of trig integration page.]
Most of the techniques you need are discussed on this page except for the sine and cosine reduction formulas. We derive them and give you practice problems on two separate pages, the sine reduction formula page and the cosine reduction formula page.

### Difference Between Trig Integration and Trig Substitution

Trig integration, covered on this page, is the evaluation of integrals that already have trig functions in the integrand.

Trig substitution is a technique that takes an integrand that most likely does NOT contain any trig functions, and uses some trig identities to introduce trig functions into the integrand. Once the integral is completely transformed, then trig integration is used to evaluate the integral. Once the evaluation is complete, another set of substitutions, based on the original ones, is done to convert the result back to the original variable.

### Trig Integration - Case List Summary (One Angle)

Trig Integration - Complete Case List Summary (One Angle)

$$\int{\sin^mx \cos^nx~dx}$$

$$n=0$$

$$\int{\sin^mx~dx}$$

use reduction formula

$$m=0$$

$$\int{\cos^mx~dx}$$

use reduction formula

odd m

$$\int{\sin^{2k+1}x\cos^nx~dx}$$

factor out $$\sin x$$, use $$\sin^2x=1-\cos^2x$$ and let $$u=\cos x$$

odd n

$$\int{\sin^mx\cos^{2k+1}x~dx}$$

factor out $$\cos x$$, use $$\cos^2x=1-\sin^2x$$ and let $$u=\sin x$$

even m and n

$$\int{\sin^{2k}x\cos^{2p}x~dx}$$

use half-angle formulas

$$\int{\sec^mx \tan^nx~dx}$$

$$n=0$$

$$\int{\sec^mx~dx}$$

use reduction formula

$$m=0$$

$$\int{\tan^nx~dx}$$

use reduction formula or use $$\sec^2x=1+\tan^2x$$, expand out
and try one of the following two cases

even m

$$\int{\sec^{2k}x\tan^nx~dx}$$

factor out $$\sec^2x$$, use $$\sec^2x=1+\tan^2x$$ and let $$u=\tan x$$

odd n

$$\int{\sec^mx\tan^{2k+1}x~dx}$$

factor out $$\sec x\tan x$$, use $$\sec^2x=1+\tan^2x$$ and let $$u=\sec x$$

none of the above 4 cases hold

convert trig functions to sine and cosine and
try the sine/cosine techniques

In order to choose the technique you need to use, you need to determine the form of the integrand and how many angles are involved.

One Angle

When all the sine and cosine terms in the integrand involve the same angle, here is what you do.

 Case 1 - one sine term only: $$\int{ \sin^n(\theta) ~d\theta}$$ Use the reduction formula found on the sine reduction formula page. Case 2 - one cosine term only: $$\int{ \cos^n(\theta) ~d\theta}$$ Use the reduction formula discussed on the cosine reduction formula page. Case 3 - odd term: $$\int{ \sin^m(\theta)~\cos^n(\theta) ~d\theta}$$ with odd m or n Factor out the odd term (if they are both odd, you can choose) and use $$\sin^2\theta + \cos^2\theta = 1$$ on the remaining term, then use substitution. Case 4 - even term: $$\int{ \sin^m(\theta)~\cos^n(\theta) ~d\theta}$$ with even m and n Use the half-angle formulas to remove the powers.

Okay, let's work some practice problems before we go on.

Instructions - - Unless otherwise instructed, evaluate the following integrals using the techniques on this page. Give all answers in exact, simplified form.

Basic Problems

$$\displaystyle{ \int{ \sin^4(2x)\cos(2x)~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sin^4(2x)\cos(2x)~dx } }$$

$$\displaystyle{ \int{ \sin^4(2x)\cos(2x)~dx } = \frac{\sin^5(2x)}{10}+C }$$

Problem Statement

$$\displaystyle{ \int{ \sin^4(2x)\cos(2x)~dx } }$$

Solution

Our first inclination might be to use the half-angle formulas for $$\sin(2x)$$ and $$\cos(2x)$$. However, on closer inspection, we can see that both of the angles are the same, i.e. both are $$2x$$. So this falls under case 3 above. The first part is done, i.e. the odd cosine is already factored out, so we are ready to use substitution. So we let $$u=\sin(2x)$$.
This gives us $$du = 2\cos(2x)dx \to du/2 = \cos(2x)dx$$

 $$\displaystyle{ \int{ \sin^4(2x)\cos(2x) ~dx } = \int{u^4 (du/2)} }$$ $$\displaystyle{\frac{1}{2}\frac{u^5}{5} + C }$$ $$\displaystyle{ \frac{u^5}{10} + C }$$ $$\displaystyle{ \frac{\sin^5(2x)}{10} + C }$$

$$\displaystyle{ \int{ \sin^4(2x)\cos(2x)~dx } = \frac{\sin^5(2x)}{10}+C }$$

$$\displaystyle{ \int{ \cos^5x\sin^5x~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \cos^5x\sin^5x~dx } }$$

Solution

For this problem, he doesn't go into a lot of detail in the integration. I would probably leave the integral as
$$\displaystyle{ \int{ (\sin^5x - 2\sin^7x + \sin^9x) } }$$ $$\cos x ~dx$$ and use the substitution $$u = \sin x$$ to get
$$\displaystyle{ \int{ u^5 - 2u^7 + u^9 ~du } = \frac{u^6}{6} - \frac{u^8}{4} + \frac{u^{10}}{10} + C }$$
The final answer follows more directly from this.

### 86 video

video by Dr Chris Tisdell

$$\displaystyle{ \int{ \cos^4x\sin^3x~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \cos^4x\sin^3x~dx } }$$

Solution

### 87 video

video by Dr Chris Tisdell

$$\displaystyle{ \int{ \sin^2x~\cos^2x~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sin^2x~\cos^2x~dx } }$$

Solution

### 94 video

video by PatrickJMT

$$\displaystyle{ \int_{0}^{ \pi/2}{\sin^2x~\cos^2x~dx } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{ \pi/2}{\sin^2x~\cos^2x~dx } }$$

Solution

### 110 video

video by Krista King Math

$$\displaystyle{ \int{ \sin^3x~\sec^2x~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sin^3x~\sec^2x~dx } }$$

Solution

### 107 video

video by MIT OCW

$$\displaystyle{\int{\sin^3x~\cos^2x~dx}}$$

Problem Statement

$$\displaystyle{\int{\sin^3x~\cos^2x~dx}}$$

Solution

### 92 video

video by PatrickJMT

$$\displaystyle{\int{\sin^3x~\cos^3x~dx}}$$

Problem Statement

$$\displaystyle{\int{\sin^3x~\cos^3x~dx}}$$

Solution

### 93 video

video by PatrickJMT

$$\displaystyle{ \int_{ \pi/2}^{\pi}{ \sin^3 \theta ~ \cos^2 \theta ~ d\theta } }$$

Problem Statement

$$\displaystyle{ \int_{ \pi/2}^{\pi}{ \sin^3 \theta ~ \cos^2 \theta ~ d\theta } }$$

Solution

### 1827 video

video by Dr Chris Tisdell

$$\int{\cos^5(\theta)~d\theta}$$

Problem Statement

$$\int{\cos^5(\theta)~d\theta}$$

$$\sin\theta-(2/3)\sin^3\theta +(1/5)\sin^5\theta+K$$

Problem Statement

$$\int{\cos^5(\theta)~d\theta}$$

Solution

At about the 4min 30sec mark in this video he jumps from $$\int{ -2\sin^2\theta\cos\theta ~d\theta }$$ to the answer $$(-2/3)\sin^3\theta$$. Later on in the video he explains how to use an identity often found in an integral table to get this.
However, we would just use integration by substitution by letting $$u=\sin\theta$$ to get this result.

### 1946 video

video by Dr Chris Tisdell

$$\sin\theta-(2/3)\sin^3\theta +(1/5)\sin^5\theta+K$$

Intermediate Problems

$$\displaystyle{ \int{ \cos^4x~\sin^2x~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \cos^4x~\sin^2x~dx } }$$

$$\displaystyle{\frac{1}{192}\left[ 12x + 4\sin^3(2x)-3\sin(4x)\right] + C}$$

Problem Statement

$$\displaystyle{ \int{ \cos^4x~\sin^2x~dx } }$$

Solution

$$\int{\cos^4 x \sin^2 x~dx}$$

use the half-angle formulas $$\cos^2 t = [1+\cos(2t)]/2$$ and $$\sin^2 t = [1-\cos(2t)]/2$$

$$\displaystyle{\int{\left[ \frac{1+\cos(2x)}{x}\right]^2 \left[ \frac{1-\cos(2x)}{2}\right]~dx}}$$

$$\displaystyle{\frac{1}{8}\int{[1+2\cos(2x)+\cos^2(2x)][1-\cos(2x)]~dx}}$$

$$\displaystyle{\frac{1}{8}\int{1+\cos(2x)-\cos^2{2x}-\cos^3(2x)~dx}}$$

on the third term, use the half-angle formula again;
on the fourth term pull out one cosine and use the identity $$\cos^2(t)+\sin^2(t)=1$$, i.e. $$\cos^3(2x)=\cos^2(2x)\cos(2x) =$$ $$(1-\sin^2(2x))\cos(2x)$$

$$\displaystyle{\frac{1}{8}\left[ x + \frac{\sin(2x)}{2} - \int{\frac{1+\cos(4x)}{2}~dx} - \int{[1-\sin^2(2x)]\cos(2x)~dx}\right]}$$

on the last term, we will use integration by substitution as follows

$$u=\sin(2x) \to du=2\cos(2x)~dx$$

$$\int{[1-\sin^2(2x)]\cos(2x)~dx} =$$ $$\int{(1-u^2)~du/2} =$$ $$(1/2)(u-u^3/3)$$

placing this last result in the full integral, we have

$$\displaystyle{\frac{1}{8}\left[ x+\frac{\sin(2x)}{2} - \frac{x}{2} - \frac{\sin(4x)}{8} - \frac{\sin(2x)}{2} + \frac{\sin^3(2x)}{6}\right] + C}$$

simplify to get the final answer

$$\displaystyle{\frac{1}{192}\left[ 12x + 4\sin^3(2x)-3\sin(4x)\right] + C}$$

$$\displaystyle{\int_{0}^{\pi/2}{\sin^7\theta~\cos^5\theta~d\theta}}$$

Problem Statement

$$\displaystyle{\int_{0}^{\pi/2}{\sin^7\theta~\cos^5\theta~d\theta}}$$

Solution

### 114 video

video by Krista King Math

$$\displaystyle{ \int{ \cos^4x~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \cos^4x~dx } }$$

Solution

### 115 video

video by PatrickJMT

$$\displaystyle{ \int{ \sin^2(\pi x)~\cos^5(\pi x)~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sin^2(\pi x)~\cos^5(\pi x)~dx } }$$

Solution

### 120 video

video by Krista King Math

Sines/Cosines in Denominator

Up until now, we have looked at only the case when all sine and cosine terms are in the numerator. So, what do you do when they are in the denominator? We will look at the case when they are mixed, i.e. terms are in both the numerator and denominator and, secondly, when all terms are in the denominator.
First, there is no definite formula to use for every case. So, to know when to use a technique, you need lots of practice and experience. Then you get a feel for what works and what doesn't.
1. When you have sine/cosine terms in both the numerator and denominator, usually substitution will work along with the techniques already listed above. You may also need to use trig identities like $$\sin^2\theta + \cos^2\theta = 1$$ to replace terms.
2. When all the terms are in the denominator, one technique to try is replacing the one in the numerator with $$\sin^2\theta + \cos^2\theta$$ and separate the integral into two terms.
Try these ideas on these practice problems.

$$\displaystyle{ \int{ \frac{\cos(x) \sin(\csc~x)}{\sin^2(x)} ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{\cos(x) \sin(\csc~x)}{\sin^2(x)} ~dx } }$$

Solution

### 119 video

video by PatrickJMT

$$\displaystyle{ \int{ \frac{\sin^2 x}{\cos^2 x}~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{\sin^2 x}{\cos^2 x}~dx } }$$

Solution

### 2130 video

video by Michel vanBiezen

$$\displaystyle{ \int{ \frac{1}{\sin x \cos^3 x}~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{1}{\sin x \cos^3 x}~dx } }$$

Solution

### 2309 video

video by Michel vanBiezen

$$\displaystyle{ \int{ \frac{\sin^3 x}{\cos x}~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{\sin^3 x}{\cos x}~dx } }$$

Hint

1. Write the numerator as $$\sin^2 x \sin x$$.
2. Replace $$\sin^2 x$$ with $$1-\cos^2 x$$.
3. Separate into two integrals, evaluating using integration by substitution.

Problem Statement

$$\displaystyle{ \int{ \frac{\sin^3 x}{\cos x}~dx } }$$

$$-\ln\abs{\cos x} - (1/2)\sin^2 x + C$$

Problem Statement

$$\displaystyle{ \int{ \frac{\sin^3 x}{\cos x}~dx } }$$

Hint

1. Write the numerator as $$\sin^2 x \sin x$$.
2. Replace $$\sin^2 x$$ with $$1-\cos^2 x$$.
3. Separate into two integrals, evaluating using integration by substitution.

Solution

### 2315 video

video by Michel vanBiezen

$$-\ln\abs{\cos x} - (1/2)\sin^2 x + C$$

$$\displaystyle{ \int{ \frac{\sin^3x}{\cos^3x} ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{\sin^3x}{\cos^3x} ~dx } }$$

$$(1/2)\tan^2x + \ln|\cos x|+C_1$$ or $$(1/2)/\cos^2x + \ln|\cos x|+C_2$$

Problem Statement

$$\displaystyle{ \int{ \frac{\sin^3x}{\cos^3x} ~dx } }$$

Solution

This integral is solved two different ways, shown in the first two videos. In the two videos, he gets what looks like two different answers. In the third video he shows that they differ only by a constant. He does it in a unique way though, by taking the derivative of the two answers and noticing they are equal. In the fourth video, he compares actual values to notice that the two answers differ by 0.5.
After all that work, I believe it is very easy to show that the answers differ only by a constant. Here is how.
Notice that the terms that don't match are $$0.5\tan^2x$$ and $$0.5/\cos^2x$$. Well, we know that $$1/\cos x = \sec x$$. So $$1/\cos^2x = \sec^2x$$. But we have the identity $$\sec^2x = 1+\tan^2x$$. So $$0.5/\cos^2x = 0.5\sec^2x = 0.5(1+\tan^2x) = 0.5+0.5\tan^2x.$$ So we know that the two answers differ by the constant 0.5. This 0.5 is absorbed in the general constant in each answer. So technically, he should have used two different symbols for the constants.

### 2316 video

video by Michel vanBiezen

### 2316 video

video by Michel vanBiezen

### 2316 video

video by Michel vanBiezen

### 2316 video

video by Michel vanBiezen

$$(1/2)\tan^2x + \ln|\cos x|+C_1$$ or $$(1/2)/\cos^2x + \ln|\cos x|+C_2$$

$$\displaystyle{ \int{ \frac{1}{\cos^3x } ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{1}{\cos^3x } ~dx } }$$

Solution

### 2333 video

video by Michel vanBiezen

Two Angles

When you have different angles use the following formulas to simplify the integrand.

 both sine terms $$\displaystyle{ \sin(\alpha)\sin(\theta) = \frac{1}{2}\left[ \cos(\alpha - \theta) - \cos(\alpha + \theta) \right] }$$ both cosine terms $$\displaystyle{ \cos(\alpha)\cos(\theta) = \frac{1}{2}\left[ \cos(\alpha - \theta) + \cos(\alpha + \theta) \right] }$$ one sine term, one cosine term $$\displaystyle{ \sin(\alpha)\cos(\theta) = \frac{1}{2}\left[ \sin(\alpha + \theta) + \sin(\alpha - \theta) \right] }$$

Try these ideas on these practice problems.

$$\displaystyle{ \int{ \sin(4x)\cos(2x)~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sin(4x)\cos(2x)~dx } }$$

Solution

### 85 video

video by Dr Chris Tisdell

$$\displaystyle{ \int{ \sin(2x)~\cos(3x)~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sin(2x)~\cos(3x)~dx } }$$

Solution

### 99 video

video by PatrickJMT

$$\displaystyle{ \int{ \sin(2x)~\sin(3x)~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sin(2x)~\sin(3x)~dx } }$$

Solution

### 100 video

video by PatrickJMT

$$\displaystyle{ \int{ \sin(8x)~\cos(5x)~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sin(8x)~\cos(5x)~dx } }$$

Solution

### 413 video

video by Krista King Math

$$\displaystyle{ \int{ \sin(3x)\sin(6x)~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sin(3x)\sin(6x)~dx } }$$

Solution

### 441 video

video by Krista King Math

$$\displaystyle{ \int{ \sin x~\cos(2x)~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sin x~\cos(2x)~dx } }$$

Solution

### 108 video

video by MIT OCW

$$\displaystyle{ \int{ \cos(4\pi x)\cos(\pi x)~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \cos(4\pi x)\cos(\pi x)~dx } }$$

Solution

### 446 video

video by Krista King Math

### trig integration 17calculus youtube playlist

You CAN Ace Calculus

 basic trig integration integration by substitution integration by parts

### Trig Identities and Formulas

basic trig identities

$$\sin^2\theta+\cos^2\theta=1$$   |   $$1+\tan^2\theta=\sec^2\theta$$

$$\displaystyle{\tan\theta=\frac{\sin\theta}{\cos\theta}}$$   |   $$\displaystyle{\cot\theta=\frac{\cos\theta}{\sin\theta}}$$

$$\displaystyle{\sec\theta=\frac{1}{\cos\theta}}$$   |   $$\displaystyle{\csc\theta=\frac{1}{\sin\theta}}$$

power reduction (half-angle) formulae

$$\displaystyle{\sin^2\theta=\frac{1-\cos(2\theta)}{2}}$$   |   $$\displaystyle{\cos^2\theta=\frac{1+\cos(2\theta)}{2}}$$

double angle formulae

$$\sin(2\theta)=2\sin\theta\cos\theta$$   |   $$\cos(2\theta)=\cos^2\theta-\sin^2\theta$$

list of trigonometric identities - wikipedia

trig sheets - pauls online notes

17calculus trig formulas - full list

### Trig Derivatives and Integrals

basic trig derivatives

$$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$

$$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$

$$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$

$$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$

$$\displaystyle{ \frac{d[\sec(t)]}{dt} = }$$ $$\sec(t)\tan(t)$$

$$\displaystyle{ \frac{d[\csc(t)]}{dt} = }$$ $$-\csc(t)\cot(t)$$

basic trig integrals

$$\int{\sin(x)~dx} = -\cos(x)+C$$

$$\int{\cos(x)~dx} = \sin(x)+C$$

$$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$

$$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$

$$\int{\sec(x)~dx} = \ln\abs{\sec(x)+\tan(x)}+C$$

$$\int{\csc(x)~dx} = -\ln\abs{\csc(x)+\cot(x)}+C$$

reduction formulae

Reduction Formulas (n is a positive integer)

$$\displaystyle{\int{\sin^n x~dx} = -\frac{\sin^{n-1}x\cos x}{n} + }$$ $$\displaystyle{ \frac{n-1}{n}\int{\sin^{n-2}x~dx} }$$

$$\displaystyle{\int{\cos^n x~dx} = \frac{\cos^{n-1}x\sin x}{n} + }$$ $$\displaystyle{ \frac{n-1}{n}\int{\cos^{n-2}x~dx}}$$

Reduction Formulas (n is an integer and $$n>1$$)

$$\displaystyle{\int{\tan^n x~dx}= \frac{\tan^{n-1}x}{n-1} - \int{\tan^{n-2}x~dx}}$$

$$\displaystyle{\int{\sec^n x~dx} = \frac{\sec^{n-2}x\tan x}{n-1} + }$$ $$\displaystyle{ \frac{n-2}{n-1}\int{\sec^{n-2}x~dx}}$$

17calculus trig formulas - full list

related topics on other pages

basic trig integration

secant-tangent trig integration

Wikipedia - List of Trig Identities

### Calculus Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

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