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Q18E
Expert-verifiedEvaluate the Integral: \(\int {{{\tan }^2}} \theta {\sec ^4}\theta d\theta \)
To evaluate the Integral\(\int {{{\tan }^2}} \theta {\sec ^4}\theta d\theta \)we will use identity \({\sec ^2}\theta = 1 + {\tan ^2}\theta \) then use the substitution method to get the result.
\(\int {{{\tan }^2}} \theta {\sec ^4}\theta d\theta = \frac{1}{3}{\tan ^3}\theta + \frac{1}{5}{\tan ^5}\theta + c\)
Let I = \(\int {{{\tan }^2}} \theta {\sec ^4}\theta d\theta \)
First, we will write \({\sec ^4}\theta = {\sec ^2}\theta {\sec ^2}\theta \) then use \({\sec ^2}\theta = 1 + {\tan ^2}\theta \)
\(\begin{aligned}{l}I &= \int {{{\tan }^2}} \theta {\sec ^2}\theta {\sec ^2}\theta d\theta \\ &= \int {{{\tan }^2}} \theta {\sec ^2}\theta (1 + {\tan ^2}\theta )d\theta \end{aligned}\)
Now we will use the substitution method
Put \(\tan \theta = x( = ){\sec ^2}\theta d\theta = dx\)
\(I = \int {({x^2}} + {x^4})dx\)
Split the integral
\(\begin{aligned}{l}I = \int {{x^2}} dx + \int {{x^4}} dx\\I = \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} + c\end{aligned}\)
\(I = \frac{{{{\tan }^3}\theta }}{3} + \frac{{{{\tan }^5}\theta }}{5} + c\)
Hence, the value of integral:
\(\int {{{\tan }^2}} \theta {\sec ^4}\theta d\theta = \frac{1}{3}{\tan ^3}\theta + \frac{1}{5}{\tan ^5}\theta + c\)
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