Question

${\int }_0^{\pi /2}\frac{d\theta }{1+\tan \theta }$ is equal to :

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

Let $I={\int }_0^{\pi /2}\frac{d\theta }{1+\tan \theta }$

$\therefore I={\int }_0^{\pi /2}\frac{d\theta }{1+\tan (\frac{\pi }{2}-\theta )}$
$={\int }_0^{\pi /2}\frac{d\theta }{1+\cot \theta }$
On adding, we get
$2I={\int }_0^{\pi /2}(\frac{1}{1+\tan \theta }+\frac{1}{1+\cot \theta })d\theta$

$={\int }_0^{\pi /2}(\frac{\cos \theta }{\cos \theta +\sin \theta }+\frac{\sin \theta }{\sin \theta +\cos \theta })d\theta$
$={\int }_0^{\pi /2}d\theta =[\theta {]}_0^{\pi /2}=\frac{\pi }{2}$

$\Rightarrow I=\frac{\pi }{4}$

Hence, ${\int }_0^{\pi /2}\frac{d\theta }{1+\tan \theta }=\frac{\pi }{4}$