Integration by parts question stuck

Question

We have to solve the following question $$\int \frac{\ln (\cos x+\sqrt{\cos 2x})}{\sin^2x} \,dx$$

In this I tried to use integration by parts as using $d(-\cot x)$

But got stuck in that .

Answer 1

Let $\displaystyle I = \int\frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx = \int \ln(\cos x+\sqrt{\cos 2x})\cdot \csc^2 xdx$

Using By parts, $\displaystyle =\ln(\cos x+\sqrt{\cos 2x})\cdot -\cot x-\int\frac{1}{\cos x+\sqrt{\cos 2x}}\cdot \bigg(-\sin x-\frac{\sin 2x}{\sqrt{\cos 2x}}\bigg)\cdot -\cot xdx$

$\displaystyle =-\cot x\cdot \ln(\cos x+\sqrt{\cos 2x})-\int\frac{\cos x-\sqrt{\cos 2x}}{\sin^2 x}\cdot \bigg(\sin x+\frac{\sin 2x}{\sqrt{\cos 2x}}\bigg)\cdot \cot xdx$

Above we have used $\left(\cos x+\sqrt{\cos 2x}\right)\cdot \left(\cos x+\sqrt{\cos 2x}\right)=\sin^2 x$

Now Let $\displaystyle J = \int\frac{\left(\cos x-\sqrt{\cos 2x}\right)}{\sin^2 x}\cdot \bigg(\sin x+\frac{\sin 2x}{\sqrt{\cos 2x}}\bigg)\cdot \cot xdx$

after simplifying

$\displaystyle J =- \int\frac{\cos x\cdot \cos 2x-2\cos^3 x}{\sin^2 x\sqrt{\cos 2x}}$

$\displaystyle = -\int\frac{\sin 2x \cdot \frac{\sin x}{\sqrt{\cos 2x}}+\cos x\cdot \sqrt{\cos 2x}}{\sin^2 x}dx = -\int \bigg(\frac{\sqrt{\cos 2x}}{\sin x}\bigg)'dx$

So $\displaystyle J = -\frac{\sqrt{\cos 2x}}{\sin x}$

So $\displaystyle I = -\cot x\cdot \ln(\cos x+\sqrt{\cos 2x})-\frac{\sqrt{\cos 2x}}{\sin x}+\mathcal{C}$