Question

Solve

$\underset{\pi /4}{\overset{\pi /2}{\int }}\frac{\cos \theta }{{\left[\cos \frac{\theta }{2}+\sin \frac{\theta }{2}\right]}^3}d\theta$

Answer 1

Let $I={\int }_{\pi /4}^{\pi /2}\frac{\cos \theta }{{\left(\cos \frac{\theta }{2}+\sin \frac{\theta }{2}\right)}^3}d\theta$

$={\int }_{\pi /4}^{\pi /2}\frac{\cos \theta /2-\sin \theta /2}{{\left(\cos \frac{\theta }{2}+\sin \frac{\theta }{2}\right)}^2}d\theta$

Let, $I^{\prime }=\int \frac{\cos \theta /2-\sin \theta /2}{(\cos \theta /2+\sin \theta /2)}d\theta$

Let, $\cos \frac{\theta }{2}+\sin \frac{\theta }{2}=x$

$d\theta (\frac{\cos \theta /2}{2}-\frac{\sin \theta /2}{2})=dx$

$\therefore I^{\prime }=2\int \frac{dx}{x^2}=\frac{-2}{x}+C$

Now $I={\int }_{\pi /4}^{\pi /2}{I}^{\prime }$

$={\left[\frac{-2}{(\cos \theta /2+\sin \theta /2)}\right]}_{\pi /4}^{\pi /2}$

$={\left(\frac{2}{\sqrt{1+\sin \theta }}\right)}_{\pi /4}^{\pi /2}$

$I=\sqrt{2}-\frac{2}{\sqrt{1+\frac{1}{\sqrt{2}}}}$