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) \[-\frac{2}{7}\left( 1-\frac{1}{8\sqrt{2}} \right)\]
Correct Answer: C
Solution :
\[\int_{0}^{\pi /2}{{{\cos }^{5}}\left( \frac{x}{2} \right)\sin x\,\,dx}\] \[=\int_{0}^{\pi /2}{{{\cos }^{5}}\left( \frac{x}{2} \right)\cdot 2\sin \left( \frac{x}{2} \right)\cos \left( \frac{x}{2} \right)dx}\] \[=2\int_{0}^{\pi /2}{\sin \left( \frac{x}{2} \right){{\cos }^{6}}\left( \frac{x}{2} \right)dx}\] 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 \[2\int_{0}^{\pi /2}{{{t}^{6}}(-2)dt}\] \[=-4\int_{0}^{\pi /2}{{{t}^{6}}\,\,dt}\] \[=-4\left[ \frac{{{t}^{7}}}{7} \right]_{0}^{\pi /2}\] \[=-\frac{4}{7}\left[ {{\cos }^{7}}\left( \frac{x}{2} \right) \right]_{0}^{\pi /2}\] \[=\frac{-4}{7}\left[ {{\cos }^{7}}\left( \frac{\pi }{4} \right)-{{\cos }^{7}}(0) \right]\] \[=\frac{-4}{7}\left[ {{\left( \frac{1}{\sqrt{2}} \right)}^{7}}-1 \right]\] \[=\frac{4}{7}\left[ 1-\frac{1}{8\sqrt{2}} \right]\]
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