Question

${\int }_0^{\frac{\pi }{2}}\frac{d\theta }{1+tan\theta }=$ [Roorkee 1980; MP PET 1996; DCE 1999]

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

Answer 1

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

$I={\int }_0^{\frac{\pi }{2}}\frac{d\theta }{1+tan\theta }={\int }_0^{\frac{\pi }{2}}\frac{d\theta }{1+tan(\frac{\pi }{2}-\theta )}={\int }_0^{\frac{\pi }{2}}\frac{d\theta }{1+cot\theta }$
$Onadding,2I={\int }_0^{\frac{\pi }{2}}(\frac{1}{1+tan\theta }+\frac{1}{1+cot\theta })d\theta$
$={\int }_0^{\frac{\pi }{2}}d\theta =[\theta {]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}$