Question

${\int }_0^{\pi /2}\frac{cos\theta -sin\theta }{(1+cos\theta )(1+sin\theta )}d\theta$ equals -

• A. $co{s}^{-1}\left(\frac{1}{\sqrt{2}}\right)$
• B. $co{s}^{-1}\left(0\right)$
• C. $co{s}^{-1}\left(1\right)$
• D. $co{s}^{-1}(-1)$

Answer 1

The correct option is C $co{s}^{-1}\left(1\right)$

Let $I={\int }_0^{\pi /2}\frac{\cos \theta -\sin \theta }{(1+\cos \theta )(1+\sin \theta )}d\theta ={\int }_0^{\pi /2}\frac{\cos (\pi /2-\theta )-\sin (\pi /2-\theta )}{(1+\cos (\pi /2-\theta \left)\right)(1+\sin (\pi /2-\theta \left)\right)}d\theta =-\int \frac{\cos \theta -\sin \theta }{(1+\cos \theta )(1+\sin \theta )}d\theta =-I$

$2I=0⟹I=0={\text{cos}}^{-1}\left(1\right)$