Finding $\int \ln (\cos x) \sin^{-2} x \, dx$

Question

I need to find $\int \ln(\cos x) \sin^{-2} x \, dx$.

I integrated by parts ($\ln (\cos x) = u$ and $\sin x\,dx = dv$) and got $-(\cos x) \ln(\cos x)+\cos x+c$. Is that correct?

I just need to know if the solution is correct?

Answer 1

Not correct as

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

$$=\ln (\cos x)\cdot\int\csc^2xdx-\int\left(\frac{d\ \ln (\cos x)}{dx}\cdot\int\csc^2xdx\right)dx$$

$$=\ln (\cos x)\cdot(-\cot x)-\int\left(\frac{-\sin x}{\cos x }\cdot(-\cot x)\right)dx$$

$$=\ln (\cos x)\cdot(-\cot x)-\int dx=\cdots $$

Answer 2

If you want to use integration by parts let dv = sin^-2(x) dx and u = ln(cosx)

Notice the integral of sin^-2(x) is -cot(x) hence v = -cot(x). and u' = -tan.

Can you finish it from there ?