Question

$\underset{0}{\overset{\pi /4}{\int }}(\cos 2x{)}^{3/2}\cdot \cos xdx=$

• A. $\frac{3\pi }{16}$
• B. $\frac{3\pi }{32}$
• C. $\frac{3\pi }{16\sqrt{2}}$
• D. $\frac{3\pi \sqrt{2}}{16}$

Answer 1

Answer: (C)

$I=\underset{0}{\overset{\pi /4}{\int }}(\cos 2x{)}^{3/2}\cos xdx$
$=\underset{0}{\overset{\pi /4}{\int }}{\left(1-2{\sin }^{2}x\right)}^{3/2}\cos xdx$
$\text{ put }\sqrt{2}\sin x=t\Rightarrow \cos xdx=dt/\sqrt{2}$
$I=\frac{1}{\sqrt{2}}\underset{0}{\overset{1}{\int }}{\left(1-t^2\right)}^{3/2}dt$
Now let $t=\sin \theta \Rightarrow dt=\cos \theta d\theta$
$I=\frac{1}{\sqrt{2}}\underset{0}{\overset{\pi /2}{\int }}{\cos }^4\theta d\theta =\frac{3\pi }{16\sqrt{2}}$