Question

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

• A. $\frac{\sqrt{\cos 2x}}{\sin x}-x+\cot x\times \ln \left[e(\cos x+\sqrt{\cos 2x})\right]+c$
• B. $\frac{\sqrt{\cos 2x}}{\sin x}+x-\cot x\times \ln \left[e(\cos x+\sqrt{\cos 2x})\right]+c$
• C. $\frac{\sqrt{\cos 2x}}{\sin x}+x+\cot x\times \ln \left[e(\cos x+\sqrt{\cos 2x})\right]+c$
• D. $\frac{\sqrt{\cos 2x}}{\sin x}-x-\cot x\times \ln \left[e(\cos x+\sqrt{\cos 2x})\right]+c$

Answer 1

The correct option is D $\frac{\sqrt{\cos 2x}}{\sin x}-x-\cot x\times \ln \left[e(\cos x+\sqrt{\cos 2x})\right]+c$

Let $I=\int \frac{\ln (\cos x+\sqrt{\cos 2x})}{{\sin }^{2}x}dx$
$=\int {\csc }^{2}x\ln \left[\sin x(\cot x+\sqrt{{\cot }^{2}x-1})\right]dx$
$=\int \ln \left(\sin x\right){\csc }^{2}xdx+\int \ln (\cot x+\sqrt{{\cot }^{2}x-1}){\csc }^{2}xdx$
$=-\ln \left(\sin x\right)\cot x-x+\int {\cot }^{2}xdx+I_1$
$=-\ln \left(\sin x\right)\cot x-\cot x-x+I_1$
where $I_1=\int \ln (\cot x+\sqrt{{\cot }^{2}x-1}){\csc }^{2}xdx$
Put $\cot x=t\Rightarrow -{\csc }^{2}xdx=dt$
$\therefore I_1=-\int \ln (t+\sqrt{t^2-1})dt$
$=-\left[\ln (t+\sqrt{t^2-1})\right]t-\int t.\frac{1+\frac{2t}{2\sqrt{t^2-1}}dt}{t+\sqrt{t^2-1}}dt$
$=-t\ln (t+\sqrt{t^2-1})+\int \frac{t}{\sqrt{t^2-1}}dt$
$=-t\ln (t+\sqrt{t^2-1})+\sqrt{t^2-1}$
$=-\cot x\ln (\cot x+\frac{\sqrt{\cos 2x}}{\sin x})+\frac{\sqrt{\cos 2x}}{\sin x}$
$\therefore I=\frac{\sqrt{\cos 2x}}{\sin x}-x-\cot x-\cot x\left(\ln (\cot x+\sqrt{\cos 2x})\right)+c$