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)

$\frac{\pi }{4}$

Answer 2

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
$2l=\int_{0}^{\pi /2}{\frac{\sin x+\cos x}{\sin x+\cos x}dx=\int_{0}^{\pi /2}{1\,dx=[x]_{0}^{\pi /2}}}$
$\Rightarrow$ $2l=\frac{\pi }{2}\,\,\,\,\,\Rightarrow \,\,\,\,l=\frac{\pi }{4}$