# I want to solve this integral : \int_0^{\frac{\pi}{2}

I want to solve this integral :
$$\displaystyle{\int_{{0}}^{{{\frac{{\pi}}{{{2}}}}}}}{\cot{{x}}}{\ln{{\left({\sec{{x}}}\right)}}}{\left.{d}{x}\right.}$$
I tried the following substitution : $$\displaystyle{\ln{{\left({\sec{{x}}}\right)}}}={t}$$ which means $$\displaystyle{\left.{d}{t}\right.}={\tan{{x}}}{\left.{d}{x}\right.}$$
$$\displaystyle{I}={\int_{{0}}^{{\infty}}}{\frac{{{\cot{{x}}}}}{{{\tan{{x}}}}}}{t}{\left.{d}{t}\right.}={\int_{{0}}^{{\infty}}}{\frac{{{t}}}{{{{\tan}^{{2}}{x}}}}}{\left.{d}{t}\right.}$$
I'm really disturbed by the $$\displaystyle{{\tan}^{{2}}{x}}$$, I tried also to substitute $$\displaystyle{\sec{{x}}}={t}$$ but it's not helpful either. Any helpful approach to solve this problem ?

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Charles Benedict
Substitute $$\displaystyle{t}={{\tan}^{{2}}{x}}$$
$$\displaystyle{\int_{{0}}^{{{\frac{{\pi}}{{{2}}}}}}}{\cot{{x}}}{\ln{{\left({\sec{{x}}}\right)}}}{\left.{d}{x}\right.}={{\frac{{14}}{\int}}_{{0}}^{{1}}}{\frac{{{\ln{{\left({1}+{t}\right)}}}}}{{{t}{\left({1}+{t}\right)}}}}{\left.{d}{t}\right.}+{{\frac{{14}}{\int}}_{{{1}}}^{{\infty}}}{\frac{{{\ln{{\left({1}+{t}\right)}}}}}{{{t}{\left({1}+{t}\right)}}}}{\left.{d}{t}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{4}}}}{\int_{{0}}^{{1}}}{\frac{{{\ln{{\left({1}+{t}\right)}}}}}{{{t}}}}{\left.{d}{t}\right.}-{{\frac{{14}}{\int}}_{{0}}^{{1}}}{\frac{{{\ln{{t}}}}}{{{1}+{t}}}}{\left.{d}{t}\right.}$$
$$\displaystyle={{\frac{{12}}{\int}}_{{0}}^{{1}}}{\frac{{{\ln{{\left({1}+{t}\right)}}}}}{{{t}}}}{\left.{d}{t}\right.}={\frac{{12}}{\cdot}}{\frac{{\pi^{{2}}}}{{{12}}}}={\frac{{\pi^{{2}}}}{{{24}}}}$$
###### Not exactly what youâ€™re looking for?
Fasaniu
Another possibility, using the Basel problem formula $$\displaystyle\sum_{{{k}\geq{1}}}{k}^{{-{2}}}={\frac{{\pi^{{2}}}}{{{6}}}}$$
We have
$$\displaystyle{I}=-{\int_{{0}}^{{{\frac{{\pi}}{{{2}}}}}}}{\frac{{{\cos{{x}}}}}{{{\sin{{x}}}}}}{\ln{{\cos{{x}}}}}{\left.{d}{x}\right.}=-{{\frac{{12}}{\int}}_{{0}}^{{{\frac{{\pi}}{{{2}}}}}}}{\frac{{{\cos{{x}}}}}{{{\sin{{x}}}}}}{\ln{{\left({1}-{{\sin}^{{2}}{x}}\right)}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{12}}{\sum}}_{{{k}\geq{1}}}{\frac{{1}}{{k}}}{\int_{{0}}^{{{\frac{{\pi}}{{{2}}}}}}}{\cos{{\left({x}\right)}}}{{\sin}^{{{2}{k}-{1}}}{\left({x}\right)}}{\left.{d}{x}\right.}$$
The integral in the summation is evaluated as
$$\displaystyle{\int_{{0}}^{{{\frac{{\pi}}{{{2}}}}}}}{\cos{{\left({x}\right)}}}{{\sin}^{{{2}{k}-{1}}}{\left({x}\right)}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{2}{k}}}}{\int_{{0}}^{{{\frac{{\pi}}{{{2}}}}}}}{d}{\left({{\sin}^{{{2}{k}}}{\left({x}\right)}}\right)}{\left.{d}{x}\right.}={\frac{{{1}}}{{{2}{k}}}}$$
Hence
$$\displaystyle{I}={\frac{{14}}{\sum}}_{{{k}\geq{1}}}{\frac{{{1}}}{{{k}^{{2}}}}}={\frac{{\pi^{{2}}}}{{{24}}}}$$
Vasquez

$$\int_0^{2\pi} \cot x \ln(\sec x)dx=\frac{1}{4}\int_0^{\frac{\pi}{2}} \frac{-2 \cos x \sin x \ln(\cos^2 x)}{1-\cos^2 x}dx$$
$$\overset{\overset{t=\cos^2 x}{dt=-2\cos x \sin x dx}}{=} \frac{1}{4} \int_0^1 \frac{\ln t}{t-1}dt=\frac14 \cdot \frac{\pi^2}{6}=\frac{\pi^2}{24}$$

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