A) \[\frac{1}{16{{a}^{3}}}\left( \frac{\pi }{4}-\frac{1}{3} \right)\]
B) \[\frac{1}{16{{a}^{3}}}\left( \frac{\pi }{4}+\frac{1}{3} \right)\]
C) \[\frac{1}{16}{{a}^{3}}\left( \frac{\pi }{4}-\frac{1}{3} \right)\]
D) \[\frac{1}{16}{{a}^{3}}\left( \frac{\pi }{4}+\frac{1}{3} \right)\]
Correct Answer: A
Solution :
Put \[x=a\tan \theta \Rightarrow dx=a{{\sec }^{2}}\theta \,d\theta ,\] then we have \[I=\int_{0}^{\pi /4}{\frac{{{a}^{4}}{{\tan }^{4}}\theta .\,a{{\sec }^{2}}\theta \,d\theta }{{{a}^{8}}{{\sec }^{8}}\theta }}\] Þ \[\frac{1}{{{a}^{3}}}\int_{0}^{\pi /4}{{{\sin }^{4}}\theta }{{\cos }^{2}}\theta \,d\theta =I=\frac{1}{{{a}^{3}}}\left[ \int_{0}^{\pi /4}{({{\sin }^{4}}\theta }-{{\sin }^{6}}\theta \right]\,d\theta \] \[=\frac{1}{{{a}^{3}}}\int_{0}^{\pi /4}{\left[ \frac{{{(1-\cos 2\theta )}^{2}}}{4}-\frac{{{(1-\cos 2\theta )}^{3}}}{8} \right]\,}d\theta \] \[=\frac{1}{8{{a}^{3}}}\int_{0}^{\pi /4}{\text{ }(1+\cos 2\theta })(1+{{\cos }^{2}}2\theta -2\cos 2\theta )d\theta \] \[=\frac{1}{8{{a}^{3}}}\int_{0}^{\pi /4}{(1-\cos 2\theta -{{\cos }^{2}}2\theta +{{\cos }^{3}}2\theta )\,d\theta }\] \[=\frac{1}{32{{a}^{3}}}\int_{0}^{\pi /4}{(2-\cos 2\theta -2\cos 4\theta +\cos 6\theta )d\theta }\] \[=\frac{1}{32{{a}^{3}}}\left[ 2\theta -\frac{\sin 2\theta }{2}-\frac{\sin 4\theta }{2}+\frac{\sin 6\theta }{6} \right]_{0}^{\pi /4}\] \[=\frac{1}{16{{a}^{3}}}\left( \frac{\pi }{4}-\frac{1}{3} \right)\].
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