Integral Calculus Question 46
Question: $ \int_{{}}^{{}}{\frac{x^{3}}{\sqrt{1-x^{8}}}dx=} $
Options:
A) $ \frac{1}{2}{{\sin }^{-1}}(x^{4})+c $
B) $ \frac{1}{3}{{\sin }^{-1}}(x^{4})+c $
C) $ \frac{1}{4}{{\sin }^{-1}}(x^{4})+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{x^{3}}{\sqrt{1-x^{8}}},dx=\int_{{}}^{{}}{\frac{x^{3}}{\sqrt{1-{{(x^{4})}^{2}}}}}} $ Put $ x^{4}=t\Rightarrow 4x^{3}dx=dt, $ then it reduces to $ \frac{1}{4}\int_{{}}^{{}}{\frac{dt}{\sqrt{1-x^{2}}}=\frac{1}{4}[{{\sin }^{-1}}(t)]+c=\frac{1}{4}{{\sin }^{-1}}(x^{4})+c.} $
BETA
