Question

lf $I_1={\int }_{1/e}^{\tan x}\frac{t}{1+t^2}dt$ and $I_{2}is{\int }_{1/e}^{\cot x}\frac{dt}{t(1+t^2)}$ then the value of $I_1+I_2$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{e}{2}$
• D. $\frac{1}{2}(e+\frac{1}{e})$

Answer 1

Answer: (B)

$1$

$I_1=\frac{1}{2}{\int }_{\frac{1}{e}}^{\tan x}\frac{2t}{1+t^2}dt=\frac{1}{2}{\left[\log (1+t^2)\right]}_{\frac{1}{e}}^{\tan x}=\frac{1}{2}[\log (1+{\tan }^{2}x)-\log (1+\frac{1}{e^2})]=\frac{1}{2}[\log \left({\sec }^{2}x\right)-\frac{\log (e^2+1)}{\log \left(e^2\right)}]=\log \sec x-\frac{\log (e^2+1)}{4}{I}_2={\int }_{\frac{1}{e}}^{\cot x}\frac{dt}{t(1+t^2)}={\int }_{\frac{1}{e}}^{\cot x}\frac{dt}{t}-\frac{1}{2}{\int }_{\frac{1}{e}}^{\cot x}\frac{2t}{1+t^2}dt={\left[\log \right]}_{\frac{1}{e}}^{\cot x}-\frac{1}{2}{\left[\log (1+e^2)\right]}_{\frac{1}{e}}^{\cot x}=\log \left(\cot x\right)-\log \frac{1}{e}-\frac{1}{2}[\log co\sec x-\log (1+\frac{1}{e^2})]=\log \left(\cot x\right)+1-\log co\sec x+\frac{1}{4}\log \left(\frac{e^2+1}{4}\right)=-\log \sin x+\log \cos x+1+\log \sin x+\frac{1}{4}\log \left(\frac{e^2+1}{4}\right)[\because -\log co\sec x=\log {\left(co\sec x\right)}^{-1}=\log sinx][\because \log \cot x=\log \left(co\sec x\right)+1+\frac{1}{4}\log \left(\frac{e^2+1}{4}\right)]I_1+I_2=1+2=\log \cos x+1+\log \left(\sec x\right)=\log \cos x+1-\log \sec x=1$