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

Answer: (C)

$-\frac{1}{4}\cos [{{\tan }^{-1}}({{x}^{4}})]+c$

Answer 2

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