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### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluating the integral $$\int {{{\tan }^5}} x{\sec ^3}xdx$$

Evaluating the integral $$\int {{{\tan }^5}} x{\sec ^3}xdxjaae$$ we use the identity $${\tan ^2}x = {\sec ^2}x - 1$$

If the power of $$\tan x$$ is odd, save a factor of $$\sec x\tan x$$ and use $${\tan ^2}x = {\sec ^2}x - 1$$to express the remaining factors in term of $$\sec x$$. Then substitute $$u = \sec x$$

See the step by step solution

## Step 1: Evaluation

Evaluating the Integral

$${\tan ^2}x = {\sec ^2}x - 1$$= $$\int {{{\tan }^4}x{{\sec }^2}} x.\sec x\tan xdx$$

= $$\int {{{\left( {{{\tan }^2}x} \right)}^2}} {\sec ^2}x.\sec xtanxdx$$

Use Trigonometric Identity $${\tan ^2}x = {\sec ^2}x - 1$$

$$\therefore \int {{{\tan }^5}} x{\sec ^2}xdx = {\int {\left( {{{\sec }^2}x - 1} \right)} ^2}{\sec ^2}x.\sec x\tan xdx \to 1$$

Let $$u = \sec x$$

Differentiating with respect to $$x$$

\begin{aligned}{l}\frac{{du}}{{dx}} &= \sec x\tan x\\ \Rightarrow du &= \sec x\tan xdx\end{aligned}

Substituting $$\sec x = u$$ and $$\sec x\tan xdx = du$$ in equation 1

Then equation 1 becomes

Therefore, $$\int {{{\tan }^5}} x{\sec ^3}xdx = {\int {\left( {{u^2} - 1} \right)} ^2}{u^2}.du \to 2$$

Use Identity $${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$$

So, $${\left( {{u^2} - 1} \right)^2} = {u^4} + 1 - 2{u^2}$$

Substitute the above value in equation 2

\begin{aligned}{l}\int {{{\tan }^5}} x{\sec ^3}xdx &= \int {\left( {{u^4} + 1 - 2{u^2}} \right)} {u^2}du\\ &= \int {\left( {{u^6} + {u^2} - 2{u^4}} \right)} du\\\therefore \int {{{\tan }^5}} x{\sec ^3}xdx &= \int {{u^6}} du + \int {{u^2}du - 2\int {{u^4}du} } \\ \Rightarrow \int {{{\tan }^5}} x{\sec ^3}xdx &= \frac{{{u^{6 + 1}}}}{{6 + 1}} + \frac{{{u^{2 + 1}}}}{{2 + 1}} - 2.\frac{{{u^{4 + 1}}}}{{4 + 1}} + c\\ \Rightarrow \int {{{\tan }^5}} x{\sec ^3}xdx &= \frac{{{u^7}}}{7} + \frac{{{u^3}}}{3} - 2.\frac{{{u^5}}}{5} + c \to 3\end{aligned}

## Step 2: Substitution

Substituting the value $$u = \sec x$$in equation 3

Thus, \begin{aligned}{l}\int {{{\tan }^5}} x{\sec ^3}xdx &= \frac{{{{\left( {\sec x} \right)}^7}}}{7} + \frac{{{{\left( {\sec x} \right)}^3}}}{3} - \frac{{2{{\left( {\sec x} \right)}^5}}}{5} + c\\ \Rightarrow \int {{{\tan }^5}} x{\sec ^3}xdx = \frac{{{{\sec }^7}x}}{7} + \frac{{{{\sec }^3}x}}{3} - \frac{{2{{\sec }^5}x}}{5} + c\end{aligned}

Hence the value of $$\int {{{\tan }^5}} x{\sec ^3}xdx$$

$$\frac{{{{\sec }^7}x}}{7} + \frac{{{{\sec }^3}x}}{3} - \frac{{2{{\sec }^5}x}}{5} + c$$