Question

Evaluate $\int \frac{sinxdx}{1+co{s}^{2}x}$

Answer 1

$\int \frac{\sin xdx}{1+{\cos }^{2}x}$

Let $\cos x=u$

Differentiating both sides, we get

$-\sin xdx=du\Rightarrow \sin xdx=-du$

$\therefore \int \frac{\sin xdx}{1+{\cos }^{2}x}=-\int \frac{du}{1+u^2}$

$\Rightarrow \int \frac{\sin xdx}{1+{\cos }^{2}x}=-{\tan }^{-1}u+C$

As $\cos x=u$

$\therefore \int \frac{\sin xdx}{1+{\cos }^{2}x}=-{\tan }^{-1}\left(\cos x\right)+C$

Hence $\int \frac{\sin xdx}{1+{\cos }^{2}x}=-{\tan }^{-1}\left(\cos x\right)+C$.