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Expert-verified Found in: Page 326 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # Evaluate the integral$$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta }$$

Evaluating the integral $$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta }$$

We use identity $${\sec ^2}\theta = 1 + {\tan ^2}\theta$$

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

See the step by step solution

## Step 1: Evaluation

Evaluating the integral$$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta }$$

\begin{aligned}{l} \Rightarrow I &= \int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^4}\theta {{\sec }^2}\theta {{\sec }^2}\theta d\theta } \\ \Rightarrow I &= \int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^4}\theta (1 + {{\tan }^2}\theta ){{\sec }^2}\theta d\theta \to 1} \end{aligned}

Let $$u = \tan \theta$$

Differentiating with respect to$$\theta$$

\begin{aligned}{l}\frac{{du}}{{d\theta }} &= {\sec ^2}\theta \\ \Rightarrow du &= {\sec ^2}\theta d\theta \end{aligned}

## Step 2: Integration Using Steps

Changing limits from $$\theta$$to $$u$$

If $$\theta = 0 \Rightarrow u = \tan o \to u = o$$

If $$\theta = \frac{\pi }{4} \Rightarrow u = \tan \frac{\pi }{4} \to u = 1$$

Upper Limit = 1 and Lower Limit = $$0$$

## Step 3: Calculation

Substituting $$\tan \theta = u$$ and $${\sec ^2}\theta d\theta = du$$ in equation 1

And changing limits from $$0$$to $$\frac{\pi }{4}$$à$$0$$to $$1$$

Then equation 1 becomes

\begin{aligned}{l} \Rightarrow I &= \int\limits_0^1 {{u^4}(1 + {u^2})du} \\ \Rightarrow I &= \int\limits_{}^1 {} \end{aligned} \begin{aligned}{l}I &= \int\limits_0^1 {{u^4}\left( {1 + {u^2}} \right)} du\\ \Rightarrow I &= \int\limits_0^1 {{u^4}du + \int\limits_0^1 {{u^6}du} } \end{aligned}

We know that $$\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}}$$

\begin{aligned}{l}I &= \left[ {\frac{{{u^{4 + 1}}}}{{4 + 1}}} \right]_0^1 + \left[ {\frac{{{u^{6 + 1}}}}{{6 + 1}}} \right]_0^1\\ &= \left[ {\frac{{{u^5}}}{5}} \right]_0^1 + \left[ {\frac{{{u^7}}}{7}} \right]_0^1\\ &= \frac{1}{5}\left( {1 - 0} \right) + \frac{1}{7}\left( {1 - 0} \right)\\ \Rightarrow I &= \frac{1}{5} + \frac{1}{7}\\ \Rightarrow I &= \frac{{7 + 5}}{{35}}\\ \Rightarrow I &= \frac{{12}}{{35}}\end{aligned}

## Step 4: Finalizing the equation

Replacing 1 by $$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta }$$

$$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta }$$$$= \frac{{12}}{{35}}$$

Hence, the value of Integral is

$$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta }$$$$= \frac{{12}}{{35}}$$ ### Want to see more solutions like these? 