Question

Evaluate $\int \frac{\ln (1+{\sin }^{2}x)}{{\cos }^{2}x}dx$

Answer 1

$I=\int \left(\frac{ln(1+{\sin }^{2}x)}{{\cos }^{2}x}\right)dx$

$=\int ({\sec }^{2}x.ln(1+{\sin }^{2}x)dx$

integrating by parts, we get

$I=tanx.ln(1+{\sin }^{2}x)-\int \left(\frac{2tanxsinxcosx}{1+{\sin }^{2}x}\right)dxI=\int (tanx.(\frac{2sinxcosx}{1+{\sin }^{2}x}\left)\right)dxI=2(\int (\frac{sinx}{cosx}).(\frac{sinxcosx}{1+{\sin }^{2}x}\left)dx\right)I=2(\int (\frac{{\sin }^{2}x}{1+{\sin }^{2}x}\left)dx\right)I=2(\int (1-\left(\frac{1}{1+{\sin }^{2}x}\right)\left)dx\right)I=2x-2(\int (\frac{1}{1+{\sin }^{2}x}\left)dx\right)$

$\int \left(\frac{1}{1+{\sin }^{-1}x}\right)dx=\int \frac{\frac{1}{{\cos }^{2}x}}{\frac{1+{sin}^{2}x}{{\cos }^{2}x}}dx$

$=\int \left(\frac{{\sec }^{2}x}{{\sec }^{2}x+{\tan }^{2}x}\right)dx$

$=\int \left(\frac{{\sec }^{2}x}{1+2{\tan }^{2}x}\right)dx$

Let $tanx=\frac{1}{\sqrt{2}}\tan \left(\theta \right)\Rightarrow {\sec }^{2}xdx=\left(\frac{1}{\sqrt{2}}\right){\sec }^2\left(\theta \right)d\left(\theta \right)$

$\therefore \int \left(\frac{{\sec }^{2}x}{1+2{\tan }^{2}x}\right)dx=\left(\frac{1}{\sqrt{2}}\right)(\int (\frac{{\sec }^2\theta }{1+2{\tan }^2\theta }d\theta =\frac{1}{\sqrt{2}}\int \frac{{\sec }^2\theta }{{\sec }^2\theta }d\theta$

$=\int \left(\frac{1}{\sqrt{2}}d\right(\theta )=\frac{\theta }{\sqrt{2}}$

$\frac{\theta }{\sqrt{2}}=\frac{{\tan }^{-1}\left(\sqrt{2}tanx\right)}{\sqrt{2}}$

Hence the correct answer is :-

$I=tanx.ln(1+{\sin }^{2}x)-2x+\sqrt{2}{\tan }^{-1}\left(\sqrt{2}tanx\right)+C.$