Integral exercise asked by Ajshe Zogjani in the math group

Exercise:

Integrate, $\int{\frac{\cos x}{\ln \left( \sin x \right)}dx}$

Solution: Let $u=\sin x\Leftrightarrow du=\cos x\,dx\Leftrightarrow dx=\frac{du}{\cos x}$

So $\int{\frac{\cos x}{\ln \left( \sin x \right)}dx}=\int{\frac{\cos x}{\ln u}\times \frac{du}{\cos x}=\int{\frac{du}{\ln u}}}$

Now Take $w=\ln u\Rightarrow {{e}^{w}}=u\Leftrightarrow {{e}^{w}}dw=du$

So $\int{\frac{du}{\ln u}=\int{\frac{{{e}^{w}}}{w}dw}}$  it’s clear that ${{e}^{w}}=1!+\frac{{{w}^{2}}}{2!}+\frac{{{w}^{3}}}{3!}+....=\sum\limits_{i=0}^{\infty }{\frac{{{w}^{i}}}{i!}}$

Hence $\frac{{{e}^{w}}}{w}={{e}^{w}}{{w}^{-1}}={{w}^{-1}}\sum\limits_{i=0}^{\infty }{\frac{{{w}^{i}}}{i!}=\sum\limits_{i=0}^{\infty }{\frac{{{w}^{i-1}}}{i!}}}=\frac{1}{w}+\sum\limits_{i=1}^{\infty }{\frac{{{w}^{i}}}{\left( i+1 \right)!}}$

Thus $\int{\frac{{{e}^{w}}}{w}dw}=\int{\sum\limits_{i=0}^{\infty }{\frac{{{w}^{i-1}}}{i!}dw}}=\int{\frac{1}{w}dw+\int{\sum\limits_{i=1}^{\infty }{\frac{{{w}^{i}}}{\left( i+1 \right)!}dw}}}$

$\Rightarrow \int{\frac{{{e}^{w}}}{w}dw}=\ln \left| w \right|+\int{\sum\limits_{i=1}^{\infty }{\frac{{{w}^{i}}}{\left( i+1 \right)!}dw}}$  in similar argument you will get the following

$\Rightarrow \int{\frac{du}{\ln u}=\ln \left| \ln u \right|+\ln u+\sum\limits_{k=2}^{\infty }{\frac{{{\left( \ln u \right)}^{k}}}{k.k!}+c}}=\ln \left| \ln \left( \sin x \right) \right|+\ln \left( \sin x \right)+\sum\limits_{k=2}^{\infty }{\frac{{{\left( \ln \sin x \right)}^{k}}}{k.k!}}+c$

Thus $\int{\frac{\cos x}{\ln \sin x}dx=\ln \left| \ln \left( \sin x \right) \right|+\ln \left( \sin x \right)+\sum\limits_{k=2}^{\infty }{\frac{{{\left( \ln \sin x \right)}^{k}}}{k.k!}+c=\operatorname{li}\left( \sin x \right)+c}}$