Mathematics - Indefinite Integration Question with Solution | TestHub

$\int \frac{{\sec }^2\theta }{\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }+\frac{2\tan \theta }{1-{\tan }^2\theta }}d\theta$

$=\int \frac{{\sec }^2\theta \left(1-{\tan }^2\theta \right)}{{\left(1+tan\theta \right)}^2}d\theta$

$=\int \frac{{\sec }^2\theta \left(1-\tan \theta \right)}{1+\tan \theta }d\theta$

Let $\tan \theta =t\Rightarrow {\sec }^2\theta d\theta =dt$

$=\int \left(\frac{1-t}{1+t}\right)dt=\int \left(-1+\frac{2}{1+t}\right)dt$

$=-t+2\ln \left(1+t\right)+C$

$=-\tan \theta +2\ln \left(1+\tan \theta \right)+C$

$\Rightarrow \lambda =-1 \mathrm{a}\mathrm{n}\mathrm{d} f\left(\theta \right)=1+\tan \theta$