Trigonometric Integrals

Calculus 2

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Before we begin, we should ask ourselves these two questions:

What types of integrals involving trigonometric functions do we currently know how to solve?
What identities do we know that might prove useful when solving?

Overview

Trigonometric integrals, to no surprise, involve the six basic trigonometric functions:

$\sin(x) \ \ \ \ \cos(x) \ \ \ \ \tan(x) \ \ \ \ \sec(x) \ \ \ \ \csc(x) \ \ \ \ \cot(x)$

The general idea is to use trigonometric identities to transform the integral into one we know how to integrate (likely via u-substitution).

This leads us to the question: what identities do we need to know?

We are interested in using the Pythagorean, Double-Angle, and Half-Angle identities.

For example, given an integral with sines and cosines, we might replace even multiples of sine using the Pythagorean identity (i.e. $\sin^2x = 1 - \cos^2x$ , $\sin^4x=(1-\cos^2x)^2$ , etc.) or we might replace even multiples of cosine using the Pythagorean identity (i.e. $\cos^2x = 1 - \sin^2x$ , $\cos^4x=(1-\sin^2x)^2$ , etc.). We would apply the same logic to other trigonometric functions and other trigonometric identites.

If interested, the derivations for each of the identities can be found below (links still in the works):

Powers of Sine and Cosine

Let's start off with an easy example.

Example 1

Compute $\int \sin^3x \cos x \ dx$ .

View Answer

$\frac{1}{4}\sin^4x +C$ or $\frac{1}{4} \cos^4x - \frac{1}{2}\cos^2x + C$

View Solution

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We have two ways of solving this problem:

Using a simple u-substitution
Using a trigonometric identity alongside a u-substitution

Let's solve the integral both ways.

Using a simple u-substitution:

Letting $u = \sin x \Rightarrow du = \cos x dx$ :

\begin{aligned} \int \sin^3x \cos x dx &= \int u^3 du \\ &= \frac{1}{4} u^4 +C\\ &= \frac{1}{4} \sin^4x +C \end{aligned}

Using a trigonometric identity alongside a u-substitution:

In an attempt to use the Pythagorean identity $\sin^2x - \cos^2x = 1 \Rightarrow \sin^2x = 1 - \cos^2x$ , we split $\sin^3$ up and get to rewriting:

\begin{aligned} \int \sin^3x \cos x dx &= \int \sin x \sin^2x \cos x dx \\ &= \int \sin x (1-\cos^2x) \cos x dx \\ &= \int (1-\cos^2x) \cos x \sin x dx \\ &= \int (\cos x -\cos^3x) \sin x dx \\ \end{aligned}

We chose to split the $\sin^3x$ term here because (1) there wasn't enough cosine terms to do anything with and because (2) replacing two of the three sines left us with the derivative of cosine (which means u-substitution would be possible -- the derivative of what we want $u$ to be is in the problem).

From here, we should make the u-substitution $u=\cos x \Rightarrow du=-\sin x dx$ :

\begin{aligned} \int \sin^3x \cos x dx &= \int (\cos x -\cos^3x) \sin x dx \\ &= - \int (u - u^3) du\\ &= - \left( \frac{1}{2}u^2 - \frac{1}{4}u^4 \right) + C\\ &= - \left( \frac{1}{2}\cos^2x - \frac{1}{4}\cos^4x \right) + C\\ &= \frac{1}{4} \cos^4x - \frac{1}{2} \cos^2x + C \end{aligned}

Note that while the solutions look different in form, they are equivalent. To prove this, we would have to use an identity (or identities) to rewrite one equation into the form of the other. We will leave this as a Practice Problem.

Suppose that our $\cos x$ term in the previous example had a higher degree. In that case, our simple u-substitution would no longer cut it. Let's think back to the identities we mentioned earlier (as well as how we solved Example 1 using Option 2) and see if we can figure out the next example.

Example 2

Compute $\int \sin^2x \cos^3 x \ dx$ .

View Answer

$\frac{1}{3}\sin^3x - \frac{1}{5}\sin^5x +C$

View Solution

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In an attempt to use the Pythagorean identity $\sin^2x - \cos^2x = 1 \Rightarrow \cos^2x = 1 - \sin^2x$ , we split $\cos^3$ up and get to rewriting:

\begin{aligned} \int \sin^2x \cos^3x dx &= \int \sin^2 x \cos^2x \cos x dx \\ &= \int \sin^2x (1-\sin^2x) \cos x dx \\ &= \int (\sin^2x-\sin^4x) \cos x dx \\ \end{aligned}

We chose to split the $\cos^3x$ term here because (1) replacing $\sin^2x$ would have converted everything into cosines which would be a harder integral and beacause (2) replacing two of the three cosines left us with the derivative of sine (which means u-substitution would be possible -- the derivative of what we want $u$ to be is in the problem).

From here, we should make the u-substitution $u=\sin x \Rightarrow du=\cos x dx$ :

\begin{aligned} \int \sin^2x \cos^3x dx &=\int (\sin^2x-\sin^4x) \cos x dx \\ &= \int (u^2 - u^4) du\\ &= \left( \frac{1}{3}u^3 - \frac{1}{5}u^5 \right) + C\\ &= \frac{1}{3}\sin^3x - \frac{1}{5}\sin^5x + C \end{aligned}

Example 3

Compute $\int \sin^2x \ dx$ .

View Answer

$\frac{1}{2}x - \frac{1}{4}\sin(2x) +C$

View Solution

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Solution coming soon.

Look back at the past three examples. Did we notice any patterns?

For the general integral $\int \sin^m(x) \cos^n(x) \ dx$ :

\begin{aligned} \boldsymbol{m} \textbf{ is odd: } & \text{Split off } \sin x \text{, use } \sin^2 x = 1 - \cos^2 x \text{, } u= \cos x \\ \boldsymbol{m} \textbf{ is even, } \boldsymbol{n} \textbf{ is odd: } & \text{Split off } \cos x \text{, use } \cos^2 x = 1 - \sin^2 x \text{, } u= \sin x\\ \boldsymbol{m}\textbf{,} \boldsymbol{n} \textbf{ are even: } & \text{Use half-angle identities} \end{aligned}

Challenge 1

Compute $\int \sin^4x \cos^5x \ dx$ .

View Answer

$\frac{1}{5}\sin^5x - \frac{2}{7}\sin^7x + \frac{1}{9}\sin^9x +C$

View Solution

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Challenge 2

Compute $\int \sin^2x \cos^2x \ dx$ .

View Answer

$\frac{1}{8}x - \frac{1}{32}\sin(4x) +C$

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Powers of Other Trig Functions

What about our other trigonometric functions? Since our strategy of using Pythagorean identities worked well for integrals involving sine and cosine, we can expect the same strategy to apply for integrals involving the other trigonometric functions as well.

For the general integral $\int \tan^m(x) \sec^n(x) \ dx$ :

\begin{aligned} \boldsymbol{m} \textbf{ is odd: } & \text{Split off } \sec x \tan x \text{, use } \tan^2 x = \sec^2 x -1 \text{, } u= \sec x \\ \boldsymbol{n} \textbf{ is even: } & \text{Split off } \sec^2 x \text{, use } \sec^2 x = \tan^2 x + 1 \text{, } u= \tan x \end{aligned}

For the general integral $\int \cot^m(x) \csc^n(x) \ dx$ :

\begin{aligned} \boldsymbol{m} \textbf{ is odd: } & \text{Split off } \csc x \cot x \text{, use } \cot^2 x = \csc^2 x -1 \text{, } u= \csc x \\ \boldsymbol{n} \textbf{ is even: } & \text{Split off } \csc^2 x \text{, use } \csc^2 x = \cot^2 x + 1 \text{, } u= \cot x \end{aligned}

Example 4

Compute $\int \tan^6x \sec^4x \ dx$ .

View Answer

$\frac{1}{7}\tan^7x + \frac{1}{9}\tan^9x +C$

View Solution

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Example 5

Compute $\int \tan^5x \sec^7x \ dx$ .

View Answer

$\frac{1}{11}\sec^{11}x - \frac{2}{9}\sec^9x + \frac{1}{7}\sec^7x +C$

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Example 6

Compute $\int \csc^4x \cot^6x \ dx$ .

View Answer

$-\frac{1}{9}\cot^9x - \frac{1}{7}\cot^7x +C$

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Note that the example above can be solved with a simple u-substitution (if you rewrite the integrand in terms of sines and cosines).

Challenge 3

Compute $\int \tan^5x \ dx$ .

View Answer

$\frac{1}{2}x - \frac{1}{4}\sin(2x) +C$

View Solution

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Practice Problems

Coming soon!

Problem 10

Prove $\frac{1}{4} \sin^4x +C = \frac{1}{4}\cos^4x - \frac{1}{2}\cos^2x + C$ (from Example 1).

Hint

Factor and use the Pythagorean identity.

View Solution

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Integration by Parts

Trigonometric Substitution