A) \[{{a}^{5}}\left[ \frac{3\pi }{16}-1 \right]\]
B) \[{{a}^{5}}\left[ \frac{3\pi }{16}+1 \right]\]
C) \[{{a}^{5}}\left[ \frac{3\pi }{16}-\frac{1}{5} \right]\]
D) None of these
Correct Answer: C
Solution :
Put \[x=a(1-\cos 2\theta )\Rightarrow dx=2a\sin 2\theta \,d\theta \] Therefore, \[\int_{0}^{a}{x{{(2ax-{{x}^{2}})}^{3/2}}dx}\] \[=\int_{0}^{\pi /4}{2{{a}^{5}}(1-\cos 2\theta ){{\sin }^{4}}2\theta \,\,d\theta }\] Now again, put \[2\theta =\varphi \] \[={{a}^{5}}\left[ \int_{0}^{\pi /2}{{{\sin }^{4}}\varphi \,d\varphi }-\int_{0}^{\pi /2}{{{\sin }^{4}}\varphi \cos \varphi \,d\varphi } \right]\]\[={{a}^{5}}\left[ \frac{3\pi }{16}-\frac{1}{5} \right]\].
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