A) \[\frac{\pi }{32}\]
B) \[\frac{\pi }{32}{{a}^{6}}\]
C) \[\frac{\pi }{16}{{a}^{6}}\]
D) \[\frac{\pi }{8}{{a}^{6}}\]
Correct Answer: B
Solution :
Put \[x=a\sin \theta \Rightarrow dx=a\cos \theta \,\,d\theta \] Now \[\int_{0}^{a}{{{x}^{4}}\sqrt{{{a}^{2}}-{{x}^{2}}}}dx={{a}^{6}}\int_{0}^{\pi /2}{{{\sin }^{4}}\theta \cos \theta \cos \theta \,d\theta }\] \[={{a}^{6}}\int_{0}^{\pi /2}{{{\sin }^{4}}\theta {{\cos }^{2}}\theta \,d\theta }\]\[={{a}^{6}}\frac{\Gamma \left( \frac{5}{2} \right).\Gamma \left( \frac{3}{2} \right)}{2\Gamma 4}=\frac{\pi }{32}{{a}^{6}}\], (Using gamma function).
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