1.3 Trigonometric Integrals

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We compute integrals involving powers and products of trigonometric functions.

1 Trigonometric Identities

The following are the Pythagorean Trigonometric Identities (named for Pythagoras of Samos) which hold for all angles, $\theta$ , in the domains of the functions involved: $\sin ^2\theta + \cos ^2\theta = 1,$ $1 + \tan ^2\theta = \sec ^2\theta ,$ and $1 + \cot ^2\theta = \csc ^2\theta .$

Next, we have the half-angle formulas:

$\sin ^2\theta = \frac {1-\cos (2\theta )}{2},$

and

$\cos ^2\theta = \frac {1+\cos (2\theta )}{2}.$ We will find the half-angle formulas useful for integrating even powers of sine and cosine.

2 Integrating Powers of Sine and Cosine

In this section, we compute integrals of the form $\int \sin ^m x \cos ^n x \; dx$ The method we use depends on whether $m$ and $n$ are even or odd.

Case 1: $m$ or $n$ is odd

In this case, the technique involves preparing and performing a $u$ -substitution. The key is to use the Pythagorean Identity $\sin ^2 \theta + \cos ^2 \theta = 1$ to convert sines to cosines or vice-versa.

example 1 Compute $\int \sin ^4 x \cos x \, dx$ This problem is solved using the substitution $u = \sin (x)$ , so that $du = \cos (x) \; dx$ . \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*}

(problem 1a) Compute $\displaystyle {\int \sin ^6 x\cos x \, dx}$

Let $u = \answer {\sin x}$ , then $du = \answer {\cos x dx}$ .

Substituting gives $\int \sin ^6 x\cos x \, dx = \int \answer {u^6} \; du$

Finally $\int \sin ^{6} x\cos x \, dx = \answer {1/7 \sin ^7 x} + C$

(problem 1b) Compute $\displaystyle {\int \sin x \cos ^4 x \, dx}$

Let $u = \answer {\cos x}$ , then $du = \answer {-\sin x dx}$ .

Substituting gives $\int \sin x\cos ^4 x \, dx = \int \answer {-u^4} \; du$

Finally $\int \sin x\cos ^4 x \, dx = \answer {-1/5 \cos ^5 x} + C$

(problem 1c) Compute $\displaystyle {\int \sin ^{10}(3x)\cos (3x) \, dx}$

Let $u = \answer {\sin (3x)}$ , then $du = \answer {3\cos (3x) dx}$ .

Substituting gives $\int \sin ^{10}(3x)\cos (3x) \, dx = \int \answer {\frac 13 u^{10}} \; du$

Finally $\int \sin ^{10}(3x)\cos (3x) \, dx = \answer {\frac {1}{33} \sin ^{11}(3x)} + C$

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

example 2 Compute $\displaystyle {\int \sin ^4 x\cos ^3 x \; dx}$

Prepare for substitution by rewriting the odd power of cosine. $\cos ^3 x = \cos ^2 x \cos x = \left [1 - \sin ^2 x\right ] \cos x$

Rewrite the original integral. $\int \sin ^4 x\cos ^3 x \; dx = \int \sin ^4 x\left [1 - \sin ^2 x\right ] \cos (x) \; dx$ Make the substitution $u = \sin x$ , $du = \cos x \, dx$ . \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 $\displaystyle {\int \sin ^6 x\cos ^3 x \; dx}$

Rewrite the odd power of cosine as

$\cos ^3 x = \left [\sin ^2 x- 1\right ] \cos x$ $\cos ^3 x = \left [1 - \sin ^2 x\right ] \cos x$ $\cos ^3 x = \left [1 - \sin x\right ]^2 \cos x$

Next, let $u = \answer {\sin x}$ so that $du = \answer {\cos x dx}$ .

Substituting gives $\int \sin ^6 x\cos ^3 x \; dx= \int \answer {u^6(1-u^2)} \; du$

Finally $\int \sin ^6 x\cos ^3 x \; dx= \answer {\sin ^7 x/7 - \sin ^9 x/9} + C$

(problem 2b) Compute $\displaystyle {\int \sin ^3 x\cos ^2 x \; dx}$

Rewrite the odd power of sine as:

$\sin ^3 x = \left [1 - \cos x\right ]^2 \sin x$ $\sin ^3 x = \left [\cos ^2 x- 1\right ] \sin x$ $\sin ^3 x = \left [1 - \cos ^2 x\right ] \sin x$

Next, let $u = \answer {\cos x}$ so that $du = \answer {-\sin x dx}$ .

Substituting gives $\int \sin ^3 x\cos ^2 x \; dx= \int \answer {-u^2(1-u^2)} \; du.$

Finally $\int \sin ^3 x\cos ^2 x \; dx= \answer {\cos ^5 x/5 - \cos ^3 x/3} + C.$

(problem 2c) Compute $\displaystyle {\int 3\sin ^8(2x)\cos ^3(2x) \; dx}$

Rewrite the odd power of cosine as:

$\cos ^3(2x) = \left [1 - \sin ^2(2x)\right ] \cos (2x)$ $\cos ^3(2x) = \left [\sin ^2(2x)- 1\right ] \cos (2x)$ $\cos ^3(2x) = \left [1 - \sin (2x)\right ]^2 \cos (2x)$

Next, let $u = \answer {\sin (2x)}$ so that $du = \answer {2\cos (2x) dx}$ .

Substituting gives $\int 3\sin ^8(2x)\cos ^3(2x) \; dx= \int \answer {3/2 u^8(1-u^2)} \; du$

Finally $\int 3\sin ^8(2x)\cos ^3(2x) \; dx= \answer {\frac 16 \sin ^9(2x)- \frac {3}{22} \sin ^{11}(2x)} + C$

(problem 2d) Compute $\displaystyle {\int \cos ^3(4x) \; dx}$

Rewrite the odd power of cosine as:

$\cos ^3(4x) = \left [\sin ^2(4x)- 1\right ] \cos (4x)$ $\cos ^3(4x) = \left [1 - \sin ^2(4x)\right ] \cos (4x)$ $\cos ^3(4x) = \left [1 - \sin (4x)\right ]^2 \cos (4x)$

Next, let $u = \answer {\sin (4x)}$ so that $du = \answer {4\cos (4x) dx}$ .

Substituting gives $\int \cos ^3(4x) \; dx= \int \answer {1/4 (1-u^2)} \; du$

Finally $\int \cos ^3(4x) \; dx= \answer {\frac 14 \sin (4x) - \frac {1}{12} \sin ^3(4x) } + C$

example 3 Compute $\displaystyle {\int \sin ^8 x\cos ^5 x \; dx}$

Prepare for substitution by rewriting the odd power of cosine. $\cos ^5 x = \cos ^4 x \cos x = \left [1 - \sin ^2 x\right ]^2 \cos x$

Rewrite the original integral. $\int \sin ^8 x\cos ^5 x \; dx = \int \sin ^8 x\left [1 - \sin ^2 x\right ]^2 \cos x \; dx.$

Make the substitution $u = \sin x, \; du = \cos x \, dx$ . \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 $\displaystyle {\int \sin ^5 x\cos ^6 x \; dx}$

Rewrite the odd power of sine as:

$\sin ^5 x = \left [1 - \cos ^4 x\right ] \sin x$ $\sin ^5 x = \left [1 + \cos ^4 x\right ] \sin x$ $\sin ^5 x = \left [1 - 2\cos ^2 x + \cos ^4 x\right ] \sin x$

Next, let $u = \answer {\cos x}$ so that $du = \answer {-\sin x dx}$ .

Substituting gives $\int \sin ^5 x\cos ^6 x \; dx= \int \answer {-u^6(1-2u^2 + u^4)} \; du$

Finally $\int \sin ^5 x\cos ^6 x \; dx= \answer {-1/7 \cos ^7 x + 2/9 \cos ^9 x - 1/11 \cos ^{11} x} + C$

(problem 3b) Compute $\displaystyle {\int \sin ^2(6x)\cos ^5(6x) \; dx}$

Rewrite the odd power of cosine as:

$\cos ^5(6x) = \left [1 - 2\sin ^2(6x) + \sin ^4(6x)\right ] \cos (6x)$ $\cos ^5(6x) = \left [1 - \sin ^4(6x)\right ] \cos (6x)$ $\cos ^5(6x) = \left [1 + \sin ^4(6x)\right ] \cos (6x)$

Next, let $u = \answer {\sin (6x)}$ so that $du = \answer {6 \cos (6x) dx}$ .

Substituting gives $\int \sin ^2(6x)\cos ^5(6x) \; dx= \int \answer {1/6 u^2(1-2u^2 + u^4)} \; du$

Finally $\int \sin ^2(6x)\cos ^5(6x) \; dx= \answer {1/18 \sin ^3(6x) - 1/15 \sin ^5(6x) + 1/42 \sin ^7(6x)} + C$

(problem 3c) Compute $\displaystyle {\int \sin ^5 x \; dx}$

Rewrite the odd power of sine as:

$\sin ^5 x = \left [1 - \cos ^4 x\right ] \sin x$ $\sin ^5 x = \left [1 + \cos ^4 x\right ] \sin x$ $\sin ^5 x = \left [1 - 2\cos ^2 x + \cos ^4 x\right ] \sin x$

Next, let $u = \answer {\cos x}$ so that $du = \answer {-\sin x dx}$ .

Substituting gives $\int \sin ^5 x \; dx= \int \answer {-(1-2u^2 + u^4)} \; du$

Finally $\int \sin ^5 x \; dx= \answer {- \cos (x) + 2/3 \cos ^3(x) - 1/5 \cos ^5(x)} + C$

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

example 4 Compute $\displaystyle {\int \sin ^7 x\cos ^5 x \; dx}$

Prepare for substitution by rewriting the smaller odd power. $\cos ^5 x = \cos ^4 x \cos x = \left [1-\sin ^2 x\right ]^2 \cos x$

Rewrite the original integral. $\int \sin ^7 x\cos ^5 x \; dx = \int \sin ^7 x\left [1 - \sin ^2 x\right ]^2 \cos x \; dx$

Make the substitution $u = \sin x, \; du = \cos x \, dx$ . \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 $\displaystyle {\int \sin ^5 x\cos ^7 x \; dx}$

Which trig function should be rewritten?

$\sin ^5 x$ $\cos ^7 x$

Rewrite the odd power of sine as:

$\sin ^5 x = \left [1 - \cos ^4 x\right ] \sin x$ $\sin ^5 x = \left [\cos ^4 x- 1\right ] \sin x$ $\sin ^5 x = \left [1 - 2\cos ^2 x + \cos ^4 x\right ] \sin x$

Next, let $u = \answer {\cos x}$ so that $du = \answer {-\sin x dx}$ .

Substituting gives $\int \sin ^5 x\cos ^7 x \; dx= \int \answer {-(1-2u^2 + u^4)u^7} \; du$

Finally $\int \sin ^5 x\cos ^7 x \; dx= \answer {-\cos ^8 x/8 + \cos ^{10}(x)/5 - \cos ^{12}(x)/12} + C$

(problem 4b) Compute $\displaystyle {\int \sin ^9 x\cos ^3 x \; dx}$

Which trig function should be rewritten?

$\sin ^9 x$ $\cos ^3 x$

Rewrite the odd power of cosine as:

$\cos ^3 x = \left [\sin ^2 x- 1\right ] \cos x$ $\cos ^3 x = \left [1 - \sin ^2 x\right ] \cos x$ $\cos ^3 x = \left [1 - \sin x\right ]^2 \cos x$

Next, let $u = \answer {\sin x}$ so that $du = \answer {\cos x dx}$ .

Substituting gives $\int \sin ^9 x\cos ^3 x \; dx= \int \answer {u^9(1 - u^2)} \; du$

Finally $\int \sin ^9 x\cos ^3 x \; dx= \answer {\sin ^{10}(x)/(10) - \sin ^{12}(x)/(12) } + C$

Case 2: $m$ and $n$ are both even

In this section, we compute integrals of the form $\int \sin ^m x \cos ^n x \; dx$ where $m$ and $n$ are both even.
We will use the half-angle formulas $\sin ^2 \theta = \frac {1 - \cos 2\theta }{2} \; \text {and} \; \cos ^2 \theta = \frac {1 + \cos 2\theta }{2}$

example 5 Compute $\displaystyle {\int \sin ^2 x \; dx}$

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*}

(problem 5) Compute $\displaystyle {\int \cos ^2 x \; dx}$

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

example 6 Compute the indefinite integral: $\int \sin ^4 x \; dx.$ 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 $\displaystyle {\int \cos ^4 x \; dx}$

Use a half angle formula to rewrite: $\cos ^4 x=$

$\tfrac 14 + \tfrac 14\cos ^2(2x)$ $\tfrac 14 - \tfrac 12\cos (2x) + \tfrac 14\cos ^2(2x)$ $\tfrac 14 + \tfrac 12\cos (2x) + \tfrac 14\cos ^2(2x)$

Use a half-angle formula again to obtain, $\cos ^4 x =$

$\tfrac 18 + \tfrac 12\cos (2x) + \tfrac 18 \cos (4x)$ $\tfrac 14 + \tfrac 12\cos (2x) + \tfrac 18 \cos (4x)$ $\tfrac 38 + \tfrac 12\cos (2x) + \tfrac 18 \cos (4x)$

Finally, $\int \cos ^4 x \; dx = \answer {\frac {3x}{8} +\frac 14 \sin (2x) + \frac {1}{32}\sin (4x)} + C$

(problem 6b) Compute $\displaystyle {\int \sin ^2(3x)\cos ^2(3x) \; dx}$

Use the half-angle formulas to rewrite: $\sin ^2(3x) \cos ^2(3x)=$

$\tfrac 14 + \tfrac 14\cos ^2(6x)$ $\tfrac 14 - \tfrac 14\sin ^2(6x)$ $\tfrac 14 - \tfrac 14\cos ^2(6x)$

Use a half-angle formula again to obtain, $\sin ^2(3x) \cos ^2(3x)=$

$\tfrac 38 - \tfrac 18 \cos (12x)$ $\tfrac 18 + \tfrac 18 \cos (12x)$ $\tfrac 18 - \tfrac 18 \cos (12x)$

Finally, $\int \sin ^2(3x)\cos ^2(3x) \; dx = \answer {\frac 18 x - \frac {1}{96} \sin (12x)} + C.$

3 Integrating Powers of Secant and Tangent

In this section, we compute integrals of the form $\int \sec ^m x\tan ^n x \; dx,$ where either $m$ is even or $n$ is odd. We will use the identity $1+\tan ^2 \theta = \sec ^2 \theta$ to prepare a $u$ -substitution.

Case 1: $m$ is even

example 7 Compute $\displaystyle {\int \tan ^4 x \sec ^2 x\;dx}$

Make the substitution $u = \tan x, \; du = \sec ^2 x \, dx$ .

\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*}

(problem 7) Compute $\displaystyle {\int \tan ^{10}(3x)\sec ^2(3x) \, dx}$

Let $u = \answer {\tan (3x)}$ , then $du = \answer {3\sec ^2(3x) dx}$ .

Substituting gives $\int \tan ^{10}(3x)\sec ^2(3x) \, dx = \int \answer {\frac 13 u^{10}} \; du$

Finally $\int \tan ^{10}(3x)\sec ^2(3x) \, dx = \answer {\frac {1}{33} \tan ^{11}(3x)} + C$

example 8 Compute $\displaystyle {\int \tan ^4 x \sec ^6 x\;dx}$

Prepare for $u$ -substitution by rewriting the even power of secant. $\sec ^6 x = \sec ^4 x \sec ^2 x = \left [1+\tan ^2 x\right ]^2 \sec ^2 x$

Rewrite the original integral.

$\int \tan ^4 x \sec ^6 x\;dx = \int \tan ^4 x \left [1+\tan ^2 x\right ]^2 \sec ^2 x \; dx$

Make the substitution $u = \tan x, \; du = \sec ^2 x \, dx$ .

\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 $\displaystyle {\int \tan ^6 x\sec ^4 x \; dx}$

Rewrite the even power of secant as

$\sec ^4 x = \left [\tan ^2 x- 1\right ] \sec ^2 x$ $\sec ^4 x = \left [1 + \tan ^2 x\right ] \sec ^2 x$ $\sec ^4 x = \left [1 - \tan ^2 x\right ] \sec ^2 x$

Next, let $u = \answer {\tan x}$ so that $du = \answer {\sec ^2 x dx}$ .

Substituting gives $\int \tan ^6 x\sec ^4 x \; dx= \int \answer {u^6(1+u^2)} \; du$

Finally $\int \tan ^6 x\sec ^4 x \; dx= \answer {\tan ^7(x)/7 + \tan ^9(x)/9} + C$

(problem 8b) Compute the indefinite integral: $\int \tan ^{1/3} x \sec ^4 x\;dx =$

Rewrite the even power of secant as

$\sec ^4 x = \left [\tan ^2 x- 1\right ] \sec ^2 x$ $\sec ^4 x = \left [1 + \tan ^2 x\right ] \sec ^2 x$ $\sec ^4 x = \left [1 - \tan ^2 x\right ] \sec ^2 x$

Next, let $u = \answer {\tan x}$ so that $du = \answer {\sec ^2 x dx}$ .

Substituting gives $\int \tan ^{1/3} x\sec ^4 x \; dx= \int \answer {u^{1/3}(1+u^2)} \; du$

Finally $\int \tan ^{1/3} x\sec ^4 x \; dx= \answer {\frac 34 \tan ^{4/3}(x) + \frac {3}{10} \tan ^{10/3}(x)} + C.$

(problem 8c) Compute the indefinite integral: $\int \sec ^6(3x)\;dx$

Rewrite the even power of secant as

$\sec ^6(3x) = \left [1 + \tan ^4(3x) \right ] \sec ^2(3x)$ $\sec ^6(3x) = \left [1 - 2\tan ^2(3x) + \tan ^4(3x) \right ] \sec ^2(3x)$ $\sec ^6(3x) = \left [1 + 2\tan ^2(3x) + \tan ^4(3x) \right ] \sec ^2(3x)$

Next, let $u = \answer {\tan (3x)}$ so that $du = \answer {3\sec ^2(3x) dx}$ .

Substituting gives $\int \sec ^6(3x) \; dx= \int \answer {(1+ 2u^2 +u^4)/3} \; du$

Finally $\int \sec ^6(3x) \; dx= \answer {\frac 13 \tan (3x) + \frac 29 \tan ^3(3x) + \frac {1}{15}\tan ^5(3x)} + C.$

Case 2: $n$ is odd

example 9

Compute $\int \sec ^5 x \tan x \, dx.$ With an odd power of tangent, we prepare the substitution $u = \sec x, \; du = \sec x\tan x \, dx$ by rewriting the integral as $\int \sec ^5 x \tan x \, dx = \int \sec ^4 x \sec x \tan x \, dx.$ 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*}

(problem 9) Compute $\displaystyle {\int \sec ^3 x \tan x \, dx}$
Let $u = \answer {\sec x}$ then $du = \answer {\sec (x)\tan (x) dx}$
Substituting gives $\int \sec ^3 x \tan x \, dx = \int \answer {u^2} \, du$ Finally $\int \sec ^3 x \tan x \, dx = \answer {1/3 \sec ^3(x)} + C$

example 10 Compute the indefinite integral: $\int \tan ^5 x \sec ^3 x\;dx.$ Rewrite the integral to prepare for the substitution, $u = \sec x, \;du = \sec x \tan x \; dx$ .

$\int \tan ^5 x \sec ^3 x\;dx = \int \tan ^4 x \sec ^2 x \sec x \tan x\;dx.$

Next, convert tangents to secants, using $1+ \tan ^2 \theta = \sec ^2 \theta$

$\tan ^4 x = \left [\sec ^2 x - 1\right ]^2$

The integral becomes

$\int \tan ^4 x \sec ^2 x \sec x \tan x\;dx = \int \left [\sec ^2 x -1\right ]^2 \sec ^2 x \sec x \tan x\;dx$

Now we are ready to substitute, $u = \sec x, \; du = \sec x \tan x \, dx$

\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: $\int \tan ^3 x \sec ^5 x\;dx$ First rewrite the integral as $\int \tan ^3 x \sec ^5 x\;dx = \int \tan ^2 x \sec ^4 x \sec x \tan x \; dx$ Next, rewrite $\tan ^2 x$ as

$1+\sec ^2 x$ $1-\sec ^2 x$ $\sec ^2 x -1$

Let $u = \answer {\sec (x)}$ , then $du = \answer {\sec (x) \tan (x) \, dx}$

Substituting gives $\int \tan ^3 x \sec ^5 x\;dx = \int \answer {(u^2 - 1)u^4} \; du$

Finally $\int \tan ^3 x \sec ^5 x\;dx = \answer {\frac 17 \sec ^7(x) - \frac 15 \sec ^5(x)} + C$

(problem 10b) Compute the indefinite integral: $\int \tan ^5 x \sec ^7 x\;dx =$ First rewrite the integral as $\int \tan ^5 x \sec ^7 x\;dx = \int \tan ^4 x \sec ^6 x \sec x \tan x \; dx$ Next, rewrite $\tan ^4 x$ as

$\left [\sec ^2 x + 1\right ]^2$ $\left [\sec ^2 x -1\right ]^2$ $\left [1-\sec ^2 x\right ]^2$

Let $u = \answer {\sec x}$ , then $du = \answer {\sec (x) \tan (x) \, dx}$

Substituting gives $\int \tan ^5 x \sec ^7 x\;dx = \int \answer {(u^4 -2u^2 + 1) u^6} \; du$

Finally $\int \tan ^5 x \sec ^7 x\;dx = \answer {\frac {1}{11} \sec ^{11}(x) - \frac 29 \sec ^9(x) + \frac 17 \sec ^7(x)} + C.$

4 Odd Powers of Secant

Compute the indefinite integral: $\int \sec x \; dx.$ This integral is computed by using a $u$ -substitution after rewriting the integrand by multiplying by 1 in the following way: $\int \sec x \; dx = \int \sec x \cdot \frac {\sec x + \tan x}{\sec x + \tan x} \; dx.$ After distributing $\sec x$ we get $\int \sec x \; dx = \int \frac {\sec ^2 x + \sec x\tan x}{\sec x + \tan x} \; dx.$ Now we observe that the numerator is the derivative of the denominator which suggests the substitution $u = \sec x + \tan x$ . We have $du = [ \sec x\tan x + \sec ^2 x] \; dx$ 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 $\sec (x)$ : $\int \sec ^n x \; dx = \frac {1}{n-1}\sec ^{n-2} x\tan x + \frac {n-2}{n-1}\int \sec ^{n-2} x \; dx.$ This formula is particularly useful if the power is odd, since if $n$ is odd, $n-2$ is also odd. Therefore, if we repeat the formula enough times, we will eventually be left with $\int \sec x \, dx$ which we have just solved.

Use the above reduction of powers formula twice to compute the indefinite integral $\int \sec ^5 x \; dx.$ We have $n =5$ , and the formula gives $\int \sec ^5 x \; dx = \frac 14 \sec ^3 x\tan x + \frac 34 \int \sec ^3 x \; dx.$ We will use the reduction of powers formula once again, this time with $n=3$ : \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:

2024-09-27 13:58:25

1 Trigonometric Identities

2 Integrating Powers of Sine and Cosine

2.1 Case 1: m or n is odd

2.2 Case 2: m and n are both even

3 Integrating Powers of Secant and Tangent

3.1 Case 1: m is even

3.2 Case 2: n is odd

4 Odd Powers of Secant

Case 1: $m$ or $n$ is odd

Case 2: $m$ and $n$ are both even

Case 1: $m$ is even

Case 2: $n$ is odd