Question

If $\underset{0}{\overset{\pi /4}{\int }}(\cos 2\theta {)}^{\frac{3}{2}}\cos \theta d\theta =\frac{a\pi }{16\sqrt{b}}$, then the value of $a+b$ is:
(where $a$ and $b$ are relative prime)

Answer 1

$I=\underset{0}{\overset{\pi /4}{\int }}(1-2{\sin }^2\theta {)}^{\frac{3}{2}}\cos \theta d\theta \text{Let}\sin \alpha =\sqrt{2}\sin \theta \Rightarrow \cos \alpha d\alpha =\sqrt{2}\cos \theta d\theta \therefore I=\underset{0}{\overset{\pi /2}{\int }}({\cos }^2\alpha {)}^{\frac{3}{2}}\frac{1}{\sqrt{2}}\cos \alpha d\alpha \Rightarrow I=\frac{1}{\sqrt{2}}\underset{0}{\overset{\pi /2}{\int }}{\cos }^4\alpha d\alpha \{\because \underset{0}{\overset{\pi /2}{\int }}{\cos }^{n}xdx=\frac{(n-1)}{n}\cdot \frac{(n-3)}{(n-2)}\cdots \frac{1}{2}\cdot \frac{\pi }{2},\text{when}n\text{is even}\}\Rightarrow I=\frac{1}{\sqrt{2}}\times \frac{3}{4}\times \frac{1}{2}\times \frac{\pi }{2}=\frac{3\pi }{16\sqrt{2}}\therefore a=3,b=2\Rightarrow a+b=5$