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