Question

Evaluate: (i) $\int \frac{\cos 2x+2{\sin }^{2}x}{{\sin }^{2}x}dx$
(ii) $\int \frac{2{\cos }^{2}x-\cos 2x}{{\cos }^{2}x}dx$

Answer 1

(i)
$$\begin{gathered} \int \left(\frac{\cos 2x+2{\sin }^{2}x}{{\sin }^{2}x}\right)dx \\ =\int \left(\frac{1-2{\sin }^{2}x+2{\sin }^{2}x}{{\sin }^{2}x}\right)dx\left[\because \cos 2x=1-2{\sin }^{2}x\right] \\ =\int {\mathrm{cosec}}^{2}xdx \\ =-\cot x+C \end{gathered}$$

(ii)
$$\begin{gathered} \int \left(\frac{2{\cos }^{2}x-\cos 2x}{{\cos }^{2}x}\right)dx \\ =\int \left(\frac{2{\cos }^{2}x-\left(2{\cos }^{2}x-1\right)}{{\cos }^{2}x}\right)dx\left[\because \cos 2x=2{\cos }^{2}x-1\right] \\ =\int {\sec }^{2}xdx \\ =\tan x+C \end{gathered}$$