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) \[\frac{{{a}^{3}}}{16}\left( \frac{\pi }{4}+\frac{1}{3} \right)\]
Correct Answer: A
Solution :
Let \[I=\int_{0}^{a}{\frac{{{x}^{4}}dx}{{{({{a}^{2}}+{{x}^{2}})}^{4}}}}\] Put \[x=a\tan \theta \]\[\Rightarrow \]\[dx=a{{\sec }^{2}}\theta \,d\theta \] \[\therefore \] \[I=\int_{0}^{\pi /4}{\frac{{{a}^{4}}{{\tan }^{4}}\theta }{{{({{a}^{2}}+{{a}^{2}}{{\tan }^{2}}\theta )}^{4}}}}a{{\sec }^{2}}\theta d\theta \] \[=\frac{1}{{{a}^{3}}}\int_{0}^{\pi /4}{\frac{{{\tan }^{4}}\theta .{{\sec }^{2}}\theta }{{{\sec }^{2}}\theta }}d\theta \] \[=\frac{1}{{{a}^{3}}}\int_{0}^{\pi /4}{{{\sin }^{4}}{{\cos }^{2}}\theta d\theta }\] \[=\frac{1}{4{{a}^{3}}}\int_{0}^{\pi /4}{(4{{\sin }^{2}}\theta {{\cos }^{2}}\theta )si{{n}^{2}}\theta d\theta }\] \[=\frac{1}{4{{a}^{3}}}\int_{0}^{\pi /4}{{{\sin }^{2}}2\theta {{\sin }^{2}}\theta \,d\theta }\] \[=\frac{1}{4{{a}^{3}}}\int_{0}^{\pi /4}{\left( \frac{1-\cos 4\theta }{2} \right)\left( \frac{1-\cos 2\theta }{2} \right)d\theta }\] \[=\frac{1}{16{{a}^{3}}}\int_{0}^{\pi /4}{(1-cos2\theta -cos4\theta }\] \[+\,\cos 2\theta \,\cos 4\theta \] \[=\frac{1}{16{{a}^{3}}}\int_{0}^{\pi /4}{[1-\cos 2\theta -\cos 4\theta }\] \[\left. +\,\,\frac{1}{2}(cos6\theta \,+\,\cos 2\theta ) \right]d\theta \] \[=\frac{1}{32{{a}^{3}}}\int_{0}^{\pi /4}{(2-cos2\theta -2cos4\theta +cos6\theta )}\,d\theta \] \[=\frac{1}{32{{a}^{3}}}\left[ 2\theta -\frac{\sin 2\theta }{2}-\frac{\sin 4\theta }{2}+\frac{\sin 6\theta }{6} \right]_{0}^{\pi /4}\] \[=\frac{1}{32{{a}^{3}}}\left[ \frac{\pi }{2}-\frac{1}{2}-0-\frac{1}{6} \right]=\frac{1}{32{{a}^{3}}}\left[ \frac{\pi }{2}-\frac{4}{6} \right]\] \[=\frac{1}{16{{a}^{3}}}\left[ \frac{\pi }{4}-\frac{1}{3} \right]\]
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