Let , then .

Substituting gives

Finally

We compute integrals involving powers and products of trigonometric functions.

The following are the Pythagorean Trigonometric Identities (named for Pythagoras of Samos) which hold for all angles, , in the domains of the functions involved: and

Next, we have the half-angle formulas:

and

We will find the half-angle formulas useful for integrating even powers of sine and cosine.

In this section, we compute integrals of the form The method we use depends on whether and are even or odd.

In this case, the technique involves preparing and performing a -substitution. The key is to use the Pythagorean Identity to convert sines to cosines or vice-versa.

example 1 Compute This problem is solved using the substitution , so that . \begin{align*} \int \sin ^4 x \cos x \, dx &= \int u^4 \, du\\ &= \frac{u^5}{5} + C\\ &= \frac 15 \sin ^5 x + C \end{align*}

In the next example, we need to prepare for the -substitution.

example 2 Compute

Prepare for substitution by rewriting the odd power of cosine.

Rewrite the original integral. Make the substitution , . \begin{align*} \int \sin ^4 x\cos ^3 x \; dx &= \int u^4 (1-u^2) \; du\\ &= \int (u^4 -u^6) \; du\\ &= \frac{u^5}{5} - \frac{u^7}{7} + C \\ &= \frac 15 \sin ^5 x - \frac 17 \sin ^7 x + C \end{align*}

(problem 2a) Compute

Rewrite the odd power of cosine as

Next, let so that .

Substituting gives

Finally

(problem 2b) Compute

Rewrite the odd power of sine as:

Next, let so that .

Substituting gives

Finally

(problem 2c) Compute

Rewrite the odd power of cosine as:

Next, let so that .

Substituting gives

Finally

(problem 2d) Compute

Rewrite the odd power of cosine as:

Next, let so that .

Substituting gives

Finally

example 3 Compute

Prepare for substitution by rewriting the odd power of cosine.

Rewrite the original integral.

Make the substitution . \begin{align*} \int \sin ^8 x\cos ^5 x \; dx &= \int u^8 (1-u^2)^2 \; du\\ &= \int u^8 (1-2u^2 + u^4) \; du\\ &= \int (u^8 -2u^{10} + u^{12}) \; du\\ &= \tfrac 19 u^9 - \tfrac{2}{11}u^{11} + \tfrac{1}{13} u^{13} + C \\ &= \tfrac 19 \sin ^9 x - \tfrac{2}{11} \sin ^{11} x + \tfrac{1}{13} \sin ^{13} x + C. \end{align*}

(problem 3a) Compute

Rewrite the odd power of sine as:

Next, let so that .

Substituting gives

Finally

(problem 3b) Compute

Rewrite the odd power of cosine as:

Next, let so that .

Substituting gives

Finally

(problem 3c) Compute

Rewrite the odd power of sine as:

Next, let so that .

Substituting gives

Finally

In the next example, both powers are odd. In this case, we have to choose which one to rewrite.

example 4 Compute

Prepare for substitution by rewriting the smaller odd power.

Rewrite the original integral.

Make the substitution . \begin{align*} \int \sin ^7 x\cos ^5 x \; dx &= \int u^7 (1-u^2)^2 \; du\\ &= \int u^7(1 - 2u^2 + u^4) \; du\\ &= \int (u^7 - 2u^9 + u^{11}) \; du\\ &= \tfrac 18 u^8 - \tfrac 15 u^{10} + \tfrac{1}{12}u^{12}+ C \\ &= \tfrac 18 \sin ^8 x - \tfrac 15 \sin ^{10} x + \tfrac{1}{12} \sin ^{12} x+ C \end{align*}

(problem 4a) Compute

Which trig function should be rewritten?

Rewrite the odd power of sine as:

Next, let so that .

Substituting gives

Finally

(problem 4b) Compute

Which trig function should be rewritten?

Rewrite the odd power of cosine as:

Next, let so that .

Substituting gives

Finally

In this section, we compute integrals of the form where and are both
even.

We will use the half-angle formulas

example 5 Compute

Rewrite the integral using a half-angle formula.

\begin{align*} \int \sin ^2 x \; dx &= \int \left [\frac 12- \frac 12\cos (2x)\right ] \; dx\\ &= \frac{x}{2} - \frac{1}{4}\sin (2x) + C. \end{align*}

example 6 Compute the indefinite integral: Rewrite the integral using the half-angle
formula for sine.

\begin{align*} \int \sin ^4 x \; dx &= \int \left [\tfrac 12- \tfrac 12\cos (2x)\right ] \cdot \left [\tfrac 12 - \tfrac 12\cos (2x)\right ] \; dx\\ &= \int \left [\tfrac 14 - \tfrac 12\cos (2x) + \tfrac 14\cos ^2(2x)\right ] \; dx \\ &= \int \left ( \tfrac 14 - \tfrac 12 \cos (2x) + \tfrac 14 \left [\tfrac 12 + \tfrac 12\cos (4x)\right ]\right ) \; dx \;\; \text{(half-angle again)} \\ &= \int \left [\tfrac 14 - \tfrac 12 \cos (2x) +\tfrac 18 + \tfrac 18\cos (4x)\right ] \; dx\\ &= \int \left [\tfrac 38 - \tfrac 12 \cos (2x) + \tfrac 18\cos (4x)\right ] \; dx \;\; \text{(now we can integrate)}\\ &= \tfrac 38 x - \tfrac 14 \sin (2x) + \tfrac{1}{32}\sin (4x) + C. \end{align*}

(problem 6a) Compute

Use a half angle formula to rewrite:

Use a half-angle formula again to obtain,

Finally,

(problem 6b) Compute

Use the half-angle formulas to rewrite:

Use a half-angle formula again to obtain,

Finally,

In this section, we compute integrals of the form where either is even or is odd. We will use the identity to prepare a -substitution.

example 7 Compute

Make the substitution .

\begin{align*} \int \tan ^4 x \sec ^2 x\;dx &= \int u^4 du\\ &= \frac{u^5}{5} + C \\ &= \frac 15\tan ^5 x + C \end{align*}

example 8 Compute

Prepare for -substitution by rewriting the even power of secant.

Rewrite the original integral.

Make the substitution .

\begin{align*} \int \tan ^4 x \sec ^6 x\;dx &= \int u^4 (1+u^2)^2 du\\ &= \int u^4(1+2u^2 + u^4) \; du \\ &= \int (u^4 + 2u^6 + u^8) \; du \\ &= \tfrac 15 u^5 + \tfrac 27 u^7 + \tfrac 19 u^9 + C \\ &= \tfrac 15\tan ^5 x + \tfrac 27 \tan ^7 x + \tfrac 19 \tan ^9 x + C \end{align*}

(problem 8a) Compute

Rewrite the even power of secant as

Next, let so that .

Substituting gives

Finally

(problem 8b) Compute the indefinite integral:

Rewrite the even power of secant as

Next, let so that .

Substituting gives

Finally

(problem 8c) Compute the indefinite integral:

Rewrite the even power of secant as

Next, let so that .

Substituting gives

Finally

example 9

Compute With an odd power of tangent, we prepare the substitution by rewriting the integral as Next we perform the substitution: \begin{align*} \int \sec ^5 x \tan x \, dx &= \int u^4 \, du\\ &= \frac{u^5}{5} + C\\ &= \frac 15 \sec ^5 x + C. \end{align*}

example 10 Compute the indefinite integral: Rewrite the integral to prepare for the
substitution, .

Next, convert tangents to secants, using

The integral becomes

Now we are ready to substitute,

\begin{align*} \int \tan ^5 x \sec ^3 x\;dx &= \int \left (u^2 -1\right )^2 u^2 \; du \\ &= \int \left (u^4 - 2u^2 +1\right ) u^2 \; du \\ &= \int \left (u^6 - 2u^4 +u^2\right ) \; du \\ &= \frac{u^7}{7} - \frac{2u^5}{5} + \frac{u^3}{3} + C \\ &= \frac 17\sec ^7 x - \frac 25 \sec ^5 x + \frac 13 \sec ^3 x + C \end{align*}

(problem 10a) Compute the indefinite integral: First rewrite the integral as Next,
rewrite as

Let , then

Substituting gives

Finally

(problem 10b) Compute the indefinite integral: First rewrite the integral as Next,
rewrite as

Let , then

Substituting gives

Finally

Compute the indefinite integral: This integral is computed by using a -substitution
after rewriting the integrand by multiplying by 1 in the following way: After
distributing we get Now we observe that the numerator is the derivative of the
denominator which suggests the substitution . We have and the integral becomes \begin{align*} \int \sec x \; dx &= \int \frac{\sec ^2 x + \sec x\tan x}{\sec x + \tan x} \; dx\\ &= \int \frac{1}{u} \; du\\ &= \ln |u| + C\\ &= \ln |\sec x + \tan x|. \end{align*}

Furthermore, IBP also gave a reduction formula for higher powers of : This formula is particularly useful if the power is odd, since if is odd, is also odd. Therefore, if we repeat the formula enough times, we will eventually be left with which we have just solved.

Use the above reduction of powers formula twice to compute the indefinite integral
We have , and the formula gives We will use the reduction of powers formula once
again, this time with : \begin{align*} \int \sec ^5 x\; dx &= \frac 14 \sec ^3 x\tan x + \frac 34 \int \sec ^3 x \; dx\\ &= \frac 14 \sec ^3 x\tan x + \frac 34\left (\frac 12\sec x\tan x + \frac 12\int \sec x \; dx\right )\\ &=\frac 14 \sec ^3 x\tan x + \frac 38\sec x\tan x + \frac 38 \int \sec x \; dx \\ &=\frac 14 \sec ^3 x\tan x + \frac 38\sec x\tan x + \frac 38\ln |\sec x + \tan x| + C \end{align*}

Here are two detailed, lecture style videos on trigonometric integrals:

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