Definite-Integration Question 543
Question: $ \int_0^{a}{x{{(2ax-x^{2})}^{\frac{3}{2}}},dx=} $
Options:
A) $ a^{5}[ \frac{3\pi }{16}-1 ] $
B) $ a^{5}[ \frac{3\pi }{16}+1 ] $
C) $ a^{5}[ \frac{3\pi }{16}-\frac{1}{5} ] $
D) None of these
Show Answer
Answer:
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}{2a^{5}(1-\cos 2\theta ){{\sin }^{4}}2\theta d\theta } $ Now again, put $ 2\theta =\varphi $ $ =a^{5}[ \int_0^{\pi /2}{{{\sin }^{4}}\varphi ,d\varphi }-\int_0^{\pi /2}{{{\sin }^{4}}\varphi \cos \varphi ,d\varphi } ] $ $ =a^{5}[ \frac{3\pi }{16}-\frac{1}{5} ] $ .
BETA
