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) \[\frac{4}{7}.\left( 1-\frac{1}{8\sqrt{2}} \right)\]
Correct Answer: D
Solution :
[d] \[I=\int\limits_{0}^{\pi /2}{{{\cos }^{5}}\left( \frac{x}{2} \right)}.\sin xdx\] \[=\int\limits_{0}^{\pi /2}{{{\cos }^{5}}\left( \frac{x}{2} \right)}.2.\sin \frac{x}{2}.dx=\int\limits_{0}^{\pi /2}{2{{\cos }^{6}}\left( \frac{x}{2} \right)}.\sin \frac{x}{2}.dx\]Let \[z=cos\frac{x}{2}\] \[\Rightarrow \frac{-1}{2}.\sin \frac{x}{2}.dx=dz\] When\[x\to 0\], then \[z\to 1\]and when \[x\to \frac{\pi }{2},z\to \frac{1}{\sqrt{2}}\] \[\therefore \]Now, \[I=\int\limits_{1}^{\frac{1}{\sqrt{2}}}{2.(-2).{{z}^{6}}.dz}\] \[I=-4\int\limits_{1}^{\frac{1}{\sqrt{2}}}{{{z}^{6}}.dz}=\frac{-4}{7}.{{\left[ {{z}^{7}} \right]}_{1}}^{\frac{1}{\sqrt{2}}}\] \[=\frac{-4}{7}.\left[ {{\left( \frac{1}{\sqrt{2}} \right)}^{7}}-1 \right]=\frac{-4}{7}\left[ \frac{1}{8\sqrt{2}}-1 \right]=\frac{4}{7}\left( 1-\frac{1}{8\sqrt{2}} \right)\]Hence, option [d] is correct.
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