Question

The value of ${\int }_0^{\pi /2}\frac{dx}{1+\tan x}$ is

• A. $\frac{\pi }{2}$
• B. $0$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{8}$

Answer 1

Answer: (C)

Let $l=\int \frac{dx}{1+\tan x}$
$\Rightarrow$ $l\int \frac{\cos x}{\sin x+\cos x}dx$ .. (i)
and $l={\int }_0^{\pi /2}\frac{\cos \left(\frac{\pi }{2}-x\right)}{\sin \left(\frac{\pi }{2}-x\right)+\cos \left(\frac{\pi }{2}-x\right)}dx$
$\Rightarrow$ $l={\int }_0^{\pi /2}\frac{\sin x}{\cos x+\sin x}dx$ .. (ii)
$\left\{\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx\right\}$
On adding Eqs. (i) and (ii), we get
$2l={\int }_0^{\pi /2}\frac{\sin x+\cos x}{\sin x+\cos x}dx={\int }_0^{\pi /2}1dx=[x{]}_0^{\pi /2}$
$\Rightarrow$ $2l=\frac{\pi }{2}\Rightarrow l=\frac{\pi }{4}$