Question

The value of $\int {\sec }^3\theta d\theta$ is

• A. $\frac{1}{2}log\text{(}sec\theta +tan\theta )+\frac{1}{2}sec\theta tan\theta +c$
• B. $log(sec\theta +tan\theta )+sec\theta tan\theta +c$
• C. $log(sec\theta -tan\theta )+\frac{1}{2}sec\theta tan\theta +c$
• D. None of the above

Answer 1

Answer: (A)

Let $I=\int {\sec }^3\theta d\theta$
$=\int \underset{I}{\mathrm{sec\theta }}\underset{II}{{\sec }^2\theta }d\theta$
$=\sec \theta (\int {\sec }^2\theta d\theta )$
$-\int \left[\left\{\frac{d}{dx}(\sec \theta )\right\}\int {\sec }^2\theta d\theta \right]d\theta$
$=\sec \theta .\tan \theta -\int \sec \theta \tan \theta .\tan \theta d\theta$
$=\sec \theta .\tan \theta -\int \sec \theta {\tan }^2\theta .d\theta$
$\Rightarrow$ $I=\sec \theta \tan \theta -\int \sec \theta ({\sec }^2\theta -1)d\theta$
$\Rightarrow$ $I=\sec \theta \tan \theta -\int {\sec }^3\theta d\theta +\int \sec \theta d\theta$
$\Rightarrow$ $I=\sec \theta \tan \theta -I+\log (\sec \theta +\tan \theta )+c_1$
$\Rightarrow$ $2I=\sec \theta \tan \theta +\log (\sec \theta +\tan \theta )+c_1$
$\Rightarrow$ $I=\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}\log (\sec \theta +\tan \theta )+\frac{c_1}{2}$
Hence, $\int {\sec }^3\theta d\theta =\frac{1}{2}\sec \theta \tan \theta$
$+\frac{1}{2}\log (\sec \theta +\tan \theta )+c$
$\left(\because c_1=\frac{c}{2}\right)$