Question

Let $A={\int }_{e^{-1}}^{tanx}\frac{tdt}{t^2+1}$ and $B={\int }_{e^{-1}}^{cotx}\frac{dt}{t(1+t^2)}$ then

• A. At $x=\frac{\pi }{4},A+B=1$
• B. A + B = 1 for all x in $(0,\frac{\pi }{2})$
• C. A + B = 1 for all x in $(0,\frac{\pi }{4})$ and 2 for all x in $(\frac{\pi }{4},\pi )$
• D. A = B for all x

Answer 1

Answer: (A), (B)

The correct options are
A

At $x=\frac{\pi }{4},A+B=1$

B

A + B = 1 for all x in $(0,\frac{\pi }{2})$

Let A + B = f(x) $={\int }_{e^{-1}}^{tanx}\frac{tdt}{t^2+1}+{\int }_{e^{-1}}^{cotx}\frac{dt}{t(1+t^2)}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{tanx}{ta{n}^{2}x+1}.se{c}^{2}x+\frac{1}{cotx(1+co{t}^{2}x)}.(-cose{c}^{2}x)=0$

$\Rightarrow f\left(x\right)$ is constant

$\Rightarrow f\left(x\right)=f\left(\frac{\pi }{4}\right)$

$\Rightarrow f\left(x\right)={\int }_{e^{-1}}^1\frac{tdt}{t^2+1}+{\int }_{e^{-1}}^1\frac{dt}{t(1+t^2)}=1$

$\Rightarrow f\left(x\right)=1$ for all x in $(0,\frac{\pi }{2})$

$\therefore$ A, B are correct.