Question

If
$I_n={\int }_{{}_{\frac{\pi }{4}}}^{{}^{\frac{\pi }{2}}}\left({\cot }^{n}x\right)dx$
then prove that $I_n+I_{n+2}=\frac{1}{n+1}$

Answer 1

$I_n={\int }_{{}_{\frac{\pi }{4}}}^{{}^{\frac{\pi }{2}}}\left({\cot }^{n}x\right)dx{I}_{n+2}={\int }_{{}_{\frac{\pi }{4}}}^{{}^{\frac{\pi }{2}}}\left({\cot }^{n+2}x\right)dx\Rightarrow I_n+I_{n+2}={\int }_{{}_{\frac{\pi }{4}}}^{{}^{\frac{\pi }{2}}}\left({\cot }^{n}x\right)dx+{\int }_{{}_{\frac{\pi }{4}}}^{{}^{\frac{\pi }{2}}}\left({\cot }^{n+2}x\right)dx\Rightarrow I_n+I_{n+2}={\int }_{{}_{\frac{\pi }{4}}}^{{}^{\frac{\pi }{2}}}\left({\cot }^{n}x\right)(1+{\cot }^{2}x)dx\Rightarrow I_n+I_{n+2}={\int }_{{}_{\frac{\pi }{4}}}^{{}^{\frac{\pi }{2}}}\left({\cot }^{n}x\right)\times {\text{cosec}}^{2}xdx$
Let $\cot x=t$
Then $-\left({\text{cosec}}^{2}x\right)dx=dt$
Now replacing $\cot x\text{with}t$
We have
$I_n+I_{n+2}=-{\int }_1^0\left(t^n\right)dt\Rightarrow I_n+I_{n+2}={\int }_0^1\left(t^n\right)dt\Rightarrow I_n+I_{n+2}={\left[\frac{t^{n+1}}{n+1}\right]}_0^1=[\frac{1}{n+1}-0]\Rightarrow I_n+I_{n+2}=\frac{1}{n+1}\text{(proved)}$