Definite Integration Question 358
Question: $ \int_0^{a}{\frac{x^{4}dx}{{{(a^{2}+x^{2})}^{4}}}}= $
Options:
A) $ \frac{1}{16a^{3}}( \frac{\pi }{4}-\frac{1}{3} ) $
B) $ \frac{1}{16a^{3}}( \frac{\pi }{4}+\frac{1}{3} ) $
C) $ \frac{1}{16}a^{3}( \frac{\pi }{4}-\frac{1}{3} ) $
D) $ \frac{1}{16}a^{3}( \frac{\pi }{4}+\frac{1}{3} ) $
Show Answer
Answer:
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 }} $
therefore $ \frac{1}{a^{3}}\int_0^{\pi /4}{{{\sin }^{4}}\theta }{{\cos }^{2}}\theta d\theta =I=\frac{1}{a^{3}}[ \int_0^{\pi /4}{({{\sin }^{4}}\theta }-{{\sin }^{6}}\theta ]d\theta $
$ =\frac{1}{a^{3}}\int_0^{\pi /4}{[ \frac{{{(1-\cos 2\theta )}^{2}}}{4}-\frac{{{(1-\cos 2\theta )}^{3}}}{8} ]}d\theta $
$ =\frac{1}{8a^{3}}\int_0^{\pi /4}{\text{ }(1+\cos 2\theta })(1+{{\cos }^{2}}2\theta -2\cos 2\theta )d\theta $
$ =\frac{1}{8a^{3}}\int_0^{\pi /4}{(1-\cos 2\theta -{{\cos }^{2}}2\theta +{{\cos }^{3}}2\theta )d\theta } $
$ =\frac{1}{32a^{3}}\int_0^{\pi /4}{(2-\cos 2\theta -2\cos 4\theta +\cos 6\theta )d\theta } $
$ =\frac{1}{32a^{3}}[ 2\theta -\frac{\sin 2\theta }{2}-\frac{\sin 4\theta }{2}+\frac{\sin 6\theta }{6} ]_0^{\pi /4} $
$ =\frac{1}{16a^{3}}( \frac{\pi }{4}-\frac{1}{3} ) $ .
BETA
