I want to solve this integral :∫0π2cot⁡xln⁡(sec⁡x)dxI tried the following substitution : ln⁡(sec⁡x)=t which means dt=tan⁡xdxI=∫0∞cot⁡xtan⁡xtdt=∫0∞ttan2xdtI'm really disturbed by the tan2x, I tried also to substitute sec⁡x=t but it's not helpful either. Any helpful approach to solve this problem ?

Question

I want to solve this integral :∫0π2cot⁡xln⁡(sec⁡x)dxI tried the following substitution : ln⁡(sec⁡x)=t which means dt=tan⁡xdxI=∫0∞cot⁡xtan⁡xtdt=∫0∞ttan2xdtI&#039;m really disturbed by the tan2x, I tried also to substitute sec⁡x=t but it&#039;s not helpful either. Any helpful approach to solve this problem ?

Charles Benedict · Accepted Answer

Substitute t=tan2x  ∫0π2cot⁡xln⁡(sec⁡x)dx=14∫01ln⁡(1+t)t(1+t)dt+14∫1∞ln⁡(1+t)t(1+t)dt  =14∫01ln⁡(1+t)tdt−14∫01ln⁡t1+tdt  =12∫01ln⁡(1+t)tdt=12⋅π212=π224

Fasaniu · Accepted Answer

Another possibility, using the Basel problem formula ∑k≥1k−2=π26  We have  I=−∫0π2cos⁡xsin⁡xln⁡cos⁡xdx=−12∫0π2cos⁡xsin⁡xln⁡(1−sin2x)dx  =12∑k≥11k∫0π2cos⁡(x)sin2k−1(x)dx  The integral in the summation is evaluated as  ∫0π2cos⁡(x)sin2k−1(x)dx=12k∫0π2d(sin2k(x))dx=12k  Hence  I=14∑k≥11k2=π224

Vasquez · Accepted Answer

∫02πcot⁡xln⁡(sec⁡x)dx=14∫0π2−2cos⁡xsin⁡xln⁡(cos2⁡x)1−cos2⁡xdx =dt=−2cos⁡xsin⁡xdxt=cos2⁡x14∫01ln⁡tt−1dt=14⋅π26=π224

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

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