Integral Calculus Question 415
Question: $ \int_{{}}^{{}}{\frac{3x^{2}}{\sqrt{9-16x^{6}}}}\ dx= $
Options:
A) $ \frac{1}{4}{{\sin }^{-1}}( \frac{4x^{3}}{3} )+c $
B) $ \frac{1}{3}{{\sin }^{-1}}( \frac{4x^{3}}{3} )+c $
C) $ \frac{1}{4}{{\sin }^{-1}}x^{3}+c $
D) $ \frac{1}{3}{{\sin }^{-1}}x^{3}+c $
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Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{3x^{2}}{\sqrt{9-16x^{6}}}dx=\int_{{}}^{{}}{\frac{3x^{2}}{\sqrt{{{(3)}^{2}}-{{(4x^{3})}^{2}}}},dx}} $ Put $ 4x^{3}=t\Rightarrow 12x^{2}dx=dt, $ then it reduces to $ \frac{1}{4}\int_{{}}^{{}}{\frac{dt}{\sqrt{{{(3)}^{2}}-t^{2}}}=\frac{1}{4}.\frac{1}{1}{{\sin }^{-1}}( \frac{t}{3} )+c=\frac{1}{4}{{\sin }^{-1}}( \frac{4x^{3}}{3} )}+c $ .
BETA
