Question

$ \int{\frac{{{x}^{3}}\sin [{{\tan }^{-1}}{{(x)}^{4}}]}{1+{{x}^{8}}}}dx $ is equal to:

• A. $ \frac{1}{4}\cos [{{\tan }^{-1}}({{x}^{4}})]+c $
• B. $ \frac{1}{4}\sin [{{\tan }^{-1}}({{x}^{4}})]+c $
• C. $ -\frac{1}{4}\cos [{{\tan }^{-1}}({{x}^{4}})]+c $
• D. $ \frac{1}{4}{{\sec }^{-1}}[{{\tan }^{-1}}({{x}^{4}})]+c $

Answer 1

The Correct Option is C

Let $ I=\int{\frac{{{x}^{3}}\sin [{{\tan }^{-1}}({{x}^{4}})]dx}{1+{{x}^{8}}}} $ Let $ {{x}^{4}}=t $ $ \Rightarrow $ $ 4{{x}^{3}}dx=dt $ $ \therefore $ $ I=\int{\frac{1}{4}\frac{\sin ({{\tan }^{-1}}(t))dt}{1+{{t}^{2}}}} $ Let $ {{\tan }^{-1}}t=u $ $ \Rightarrow $ $ \frac{1}{1+{{t}^{2}}}dt=du $ $ \therefore $ $ I=\int{\frac{1}{4}\sin u\,du} $ $ =-\frac{1}{4}\cos u+c=-\frac{1}{4}\cos {{\tan }^{-1}}(t)+c $ $ =-\frac{1}{4}\cos {{\tan }^{-1}}({{x}^{4}})+c $