Question

$ \int_{0}^{\pi /2}{{{\cos }^{5}}\left( \frac{x}{2} \right)}\sin x\,\,dx $ is equal to

• A. $ \frac{2}{7}\left( 1-\frac{1}{8\sqrt{2}} \right) $
• B. $ -\frac{4}{7}\left( 1-\frac{1}{8\sqrt{2}} \right) $
• C. $ \frac{4}{7}\left( 1-\frac{1}{8\sqrt{2}} \right) $
• D. $ -\frac{2}{7}\left( 1-\frac{1}{8\sqrt{2}} \right) $

Answer 1

Answer: (C)

$ \int_{0}^{\pi /2}{{{\cos }^{5}}\left( \frac{x}{2} \right)\sin x\,\,dx} $
$ =\int_{0}^{\pi /2}{{{\cos }^{5}}\left( \frac{x}{2} \right)\cdot 2\sin \left( \frac{x}{2} \right)\cos \left( \frac{x}{2} \right)dx} $
$ =2\int_{0}^{\pi /2}{\sin \left( \frac{x}{2} \right){{\cos }^{6}}\left( \frac{x}{2} \right)dx} $
By putting $ \cos \frac{x}{2}=t $ , $ -\frac{1}{2}\sin \frac{x}{2}dx=dt $
$ \sin \frac{x}{2}dx=-2dt $
$ \sin \frac{x}{2}dx=-2dt $
The given integral becomes
$ 2\int_{0}^{\pi /2}{{{t}^{6}}(-2)dt} $
$ =-4\int_{0}^{\pi /2}{{{t}^{6}}\,\,dt} $
$ =-4\left[ \frac{{{t}^{7}}}{7} \right]_{0}^{\pi /2} $
$ =-\frac{4}{7}\left[ {{\cos }^{7}}\left( \frac{x}{2} \right) \right]_{0}^{\pi /2} $
$ =\frac{-4}{7}\left[ {{\cos }^{7}}\left( \frac{\pi }{4} \right)-{{\cos }^{7}}(0) \right] $
$ =\frac{-4}{7}\left[ {{\left( \frac{1}{\sqrt{2}} \right)}^{7}}-1 \right] $
$ =\frac{4}{7}\left[ 1-\frac{1}{8\sqrt{2}} \right] $