Question

Prove that ${\int }_{1/e}^{\tan x}\frac{t}{1+t^2}dt+{\int }_{1/e}^{\cot x}\frac{1}{t(1+t^2)}dt=1$.

Answer 1

To prove ${\int }_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+{\int }_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$

Let $I_1={\int }_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt{I}_2={\int }_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$

and $I=I_1+I_2$

also, ${\int }_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$

Let $t=\frac{1}{z}$

$dt=\frac{-1}{z^2}dz$

So, $I_1={\int }_e^{\tan x}\frac{-zdz}{z^2(1+\frac{1}{z^2})}={\int }_{\tan x}^e\frac{zdz}{1+z^2}$

So, $I_2={\int }_{\tan x}^e\frac{tdt}{1+t^2}$

So, $I=I_1+I_2$

$I={\int }_{\frac{1}{e}}^e\frac{tdt}{1+t^2}=\frac{1}{2}{\int }_{\frac{1}{e}}^e\frac{2t}{1+t^2}dtI=\frac{1}{2}{\log (1+t^2)}_{\frac{1}{e}}^{e}I=\frac{1}{2}\log (1+e^2)-\frac{1}{2}\log (1+\frac{1}{e^2})I=\frac{1}{2}\left[\log \left(\frac{1+e^2}{1+\frac{1}{e^2}}\right)\right]I=\frac{1}{2}\log e^{2}I=\frac{1}{2}\times 2I=1$