• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q26E

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$$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt}$$

Evaluating the Integral $$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt}$$, we use Trigonometric Identity, $${\tan ^2}x = {\sec ^2}x - 1$$ then substitution method to integrate it.

See the step by step solution

## Step 1: Evaluation

Evaluating the integration $$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt}$$

Let $$I = \int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt}$$

$$I = \int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^4}t \times {{\tan }^2}} tdt$$

We use $${\tan ^2}t = {\sec ^2}t - 1$$

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

## Step 2: Using Limits

Finding Integral $$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^2}} t{\sec ^2}tdt$$

Let $$u = \tan t$$

\begin{aligned}{l}\frac{{du}}{{dt}} &= {\sec ^2}t\\ \Rightarrow du = {\sec ^2}tdt\end{aligned}

If $$t = 0 \Rightarrow u = \tan 0 \to 0$$

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

Therefore Lower Limit = 0 and Upper Limit = 1

\begin{aligned}{l}\therefore \int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^2}} t{\sec ^2}tdt &= \int\limits_0^1 {{u^2}.du} \\ &= \left( {\frac{{{u^{2 + 1}}}}{{2 + 1}}} \right)_0^1\\ &= \left( {\frac{{{u^3}}}{3}} \right)_0^1\\ &= \frac{1}{3}\left( {1 - 0} \right)\\\therefore \int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^2}} t{\sec ^2}tdt &= \frac{1}{3} \to 2\end{aligned}

## Step 3:Calculation

Finding the Integral of $$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\sec }^2}tdt}$$

\begin{aligned}{l}\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\sec }^2}} tdt &= \left( {\tan t} \right)_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}}\\ &= \tan \frac{\pi }{4} - \tan 0\\ &= 1 - 0\\ &= 1\\\therefore \int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\sec }^2}tdt &= 1 \to 3} \end{aligned}

Finding the value of $$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {dt}$$

\begin{aligned}{l}\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {dt &= \left( t \right)_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}}} \\ &= \frac{\pi }{4} - 0\\\therefore \int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {dt &= \frac{\pi }{4}} \to 4\end{aligned}

## Step 4: Substitution

Substituting the values of equations 2, 3 and 4

\begin{aligned}{l}\therefore I &= \frac{1}{3} - 1 + \frac{\pi }{4}\\ \Rightarrow I &= \frac{{4 - 12 + 3\pi }}{{12}}\\ \Rightarrow I &= \frac{{ - 8 + 3\pi }}{{12}} \Rightarrow I &= \frac{{3\pi - 8}}{{12}}\end{aligned}

## Step 5: Replacing the equation

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

Thus, $$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt} = \frac{{3\pi - 8}}{{12}}$$

Hence the value of $$\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt}$$ is $$\frac{{3\pi - 8}}{{12}}$$