Skills Review for Trigonometric Integrals

Learning Outcomes

Apply the Pythagorean identities
Apply the product-to-sum formulas
Use substitution to evaluate indefinite integrals containing trigonometric functions

In the Trigonometric Integrals section, we will learn how to evaluate integrals that contain trigonometric functions raised to powers. Here we will review trigonometric identifies and how to use substitution to evaluate trigonometric integrals.

Apply the Pythagorean Identities

Pythagorean identities are equations involving trigonometric functions based on the properties of a right triangle.

Pythagorean Identities
${\sin }^{2}\theta +{\cos }^{2}\theta =1$	$1+{\tan }^{2}\theta ={\sec }^{2}\theta$	$1+{\cot }^{2}\theta ={\csc }^{2}\theta$

We are especially interested in the first two Pythagorean Identities for the Trigonometric Integrals section.

Notice the identity

\sin^{2} θ + \cos^{2} θ = 1

${\sin }^{2}\theta +{\cos }^{2}\theta =1$ can be rearranged into the following useful alternative forms:

${\cos }^{2}\theta =1-{\sin }^{2}\theta$
${\sin }^{2}\theta =1 - {\cos }^{2}\theta$

Notice the identity

1 + \tan^{2} θ = \sec^{2} θ

$1+{\tan }^{2}\theta ={\sec }^{2}\theta$ can be rearranged into the following useful alternative forms:

${\tan }^{2}\theta = {\sec }^{2}\theta - 1$
${\sec }^{2}\theta = 1+{\tan }^{2}\theta$ (same as the original identity but sides of the equation are swapped)

Example: Using Pythagorean Identities to Rewrite Expressions

Simplify $\frac{{\sec }^{2}\theta -1}{{\sec }^{2}\theta }$ .

Show Solution

$\begin{align}\frac{{\sec }^{2}\theta -1}{{\sec }^{2}\theta }&=\frac{\left({\tan }^{2}\theta +1\right)-1}{{\sec }^{2}\theta }&& {\sec}^{2}\theta ={\tan }^{2}\theta +1 \\ &=\frac{{\tan }^{2}\theta }{{\sec }^{2}\theta } \\ &={\tan }^{2}\theta \left(\frac{1}{{\sec }^{2}\theta }\right) \\ &={\tan }^{2}\theta \left({\cos }^{2}\theta \right)&& {\cos }^{2}\theta =\frac{1}{{\sec }^{2}\theta } \\ &=\left(\frac{{\sin }^{2}\theta }{{\cos }^{2}\theta }\right)\left({\cos }^{2}\theta \right)&& {\tan}^{2}\theta =\frac{{\sin }^{2}\theta }{{\cos }^{2}\theta } \\ &=\left(\frac{{\sin }^{2}\theta }{\cancel{{\cos }^{2}\theta}}\right)\left(\cancel{{\cos }^{2}\theta} \right) \\ &={\sin }^{2}\theta \end{align}$

Example: Using Pythagorean Identities to Rewrite Expressions

Rewrite the expression ${\sin}\theta {\cos^4}\theta$ using the identity ${\cos }^{2}\theta =1 - {\sin }^{2}\theta$ .

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$\begin{align}{\sin}\theta {\cos^4}\theta&={\sin}\theta ({cos^2}\theta)^2 \\ &={\sin}\theta (1-{\sin^2}\theta)^2 && {\cos }^{2}\theta =1 - {\sin }^{2}\theta \\ &= {\sin}\theta (1-2{\sin^2}\theta +{\sin^4}\theta) \\ &={\sin}\theta-2{\sin^3}\theta+{\sin^5}\theta \end{align}$

Try It

Rewrite the expression ${\tan^4}\theta$ using the identity ${\tan }^{2}\theta = {\sec }^{2}\theta - 1$ .

Show Solution

${\sec^4}\theta-2{\sec^2}\theta+1$

Try It

Rewrite the expression ${\sin^2}\theta {\cos}\theta$ using the identity ${\sin }^{2}\theta =1 - {\cos }^{2}\theta$ .

Show Solution

${\cos}\theta-{\cos^3}\theta$

Apply the Product-to-Sum Formulas

Sometimes, we may need to express the product of cosine and sine as a sum. We can use the product-to-sum formulas, which express products of trigonometric functions as sums. Note the following product-to-sum formulas.

$\cos \alpha \cos \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)+\cos \left(\alpha +\beta \right)\right]$
$\sin \alpha \cos \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right]$
$\sin \alpha \sin \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)-\cos \left(\alpha +\beta \right)\right]$
$\cos \alpha \sin \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)-\sin \left(\alpha -\beta \right)\right]$

Example: Expanding Using a Product-To-Sum Formula

Write the following product of cosines as a sum: $2\cos \left(\frac{7x}{2}\right)\cos \left(\frac{3x}{2}\right)$ .

Show Solution

We begin by writing the formula for the product of cosines:

$\cos \alpha \cos \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)+\cos \left(\alpha +\beta \right)\right]$

We can then substitute the given angles into the formula and simplify.

$\begin{align}2\cos \left(\frac{7x}{2}\right)\cos \left(\frac{3x}{2}\right)&=\left(2\right)\left(\frac{1}{2}\right)\left[\cos \left(\frac{7x}{2}-\frac{3x}{2}\right)+\cos \left(\frac{7x}{2}+\frac{3x}{2}\right)\right] \\ &=\left[\cos \left(\frac{4x}{2}\right)+\cos \left(\frac{10x}{2}\right)\right] \\ &=\cos 2x+\cos 5x \end{align}$

Example: Expanding Using a Product-To-Sum Formula

Express the following product as a sum containing only sine or cosine and no products: $\sin \left(4\theta \right)\cos \left(2\theta \right)$ .

Show Solution

Write the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify.

$\begin{align}\sin \alpha \cos \beta &=\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right] \\ \sin \left(4\theta \right)\cos \left(2\theta \right)&=\frac{1}{2}\left[\sin \left(4\theta +2\theta \right)+\sin \left(4\theta -2\theta \right)\right] \\ &=\frac{1}{2}\left[\sin \left(6\theta \right)+\sin \left(2\theta \right)\right] \end{align}$

Try It

Use the product-to-sum formula to write the product as a sum or difference: $\cos \left(2\theta \right)\cos \left(4\theta \right)$ .

Show Solution

\frac{1}{2} (\cos 6 θ + \cos 2 θ)

$\frac{1}{2}\left(\cos 6\theta +\cos 2\theta \right)$

Try It

Use Substitution to Evaluate Indefinite Integrals Containing Trigonometric Functions

We can generalize substitution using the following steps:

Look carefully at the integrand and select an expression $g(x)$ within the integrand to set equal to $u$ . Let’s select $g(x).$ such that ${g}^{\prime }(x)$ is also part of the integrand.
Substitute $u=g(x)$ and $du={g}^{\prime }(x)dx.$ into the integral.
We should now be able to evaluate the integral with respect to $u$ . If the integral can’t be evaluated we need to go back and select a different expression to use as $u$ .
Evaluate the integral in terms of $u$ .
Write the result in terms of $x$ and the expression $g(x).$

Example: Applying Substitution to Integrals with Trigonometric Functions

Use substitution to evaluate the integral $\displaystyle\int \frac{ \sin t}{{ \cos }^{3}t}dt.$

Show Solution

We know the derivative of $\cos t$ is $\text{−} \sin t,$ so we set $u= \cos t.$ Then $du=\text{−} \sin tdt.$ Substituting into the integral, we have

$\displaystyle\int \frac{ \sin t}{{ \cos }^{3}t}dt=\text{−}\displaystyle\int \frac{du}{{u}^{3}}.$

Evaluating the integral, we get

$\begin{array}{}\\ \\ \text{−}{\displaystyle\int \frac{du}{{u}^{3}}}\hfill & =\text{−} {\displaystyle\int {u}^{-3}du}\hfill \\ & =\text{−}(-\frac{1}{2}){u}^{-2}+C.\hfill \end{array}$

Putting the answer back in terms of $t$ , we get

$\begin{array}{cc} {\displaystyle\int \frac{ \sin t}{{ \cos }^{3}t}dt}\hfill & =\frac{1}{2{u}^{2}}+C\hfill \\ \\ & =\dfrac{1}{2{ \cos }^{2}t}+C.\hfill \end{array}$

Example: Applying Substitution to Integrals with Trigonometric Functions

Use substitution to evaluate the integral $\displaystyle\int {\tan}(t) {\sec^2}(t) dt.$

Show Solution

We know the derivative of $\tan t$ is ${sec^2}t,$ so we set $u= \tan t.$ Then $du={\sec^2}t dt.$ Substituting into the integral, we have

$\displaystyle\int {\tan}(t) {\sec^2}(t) dt=\displaystyle\int u du.$

Evaluating the integral, we get

$\begin{array}{}\\ \\ {\displaystyle\int u du}\hfill & = \dfrac{u^2}{2} + C \hfill \end{array}$

Putting the answer back in terms of $t$ , we get

$\dfrac{u^2}{2} + C=\dfrac{{\tan^2}t}{2} + C$

Try It

Use substitution to evaluate the integral $\displaystyle\int \frac{ \cos t}{{ \sin }^{2}t}dt.$

Show Solution

$-\dfrac{1}{ \sin t}+C$

Try It

Use substitution to evaluate the indefinite integral $\displaystyle\int { \cos }^{3}(t) \sin (t)dt.$

Show Solution

$-\dfrac{{ \cos }^{4}t}{4}+C$

Techniques of Integration: Prerequisite Material