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) \[\frac{1}{4}{{\sec }^{-1}}[{{\tan }^{-1}}({{x}^{4}})]+c\]
E) \[-\frac{1}{4}{{\cos }^{-1}}[{{\tan }^{-1}}({{x}^{4}})]+c\]
Correct Answer: C
Solution :
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 \[{{\tan }^{-1}}t=u\] \[\Rightarrow \] \[\frac{1}{1+{{t}^{2}}}dt=du\] \[\therefore \] \[I=\int{\frac{1}{4}\sin u\,du}\] \[=-\frac{1}{4}\cos u+c=-\frac{1}{4}\cos {{\tan }^{-1}}(t)+c\] \[=-\frac{1}{4}\cos {{\tan }^{-1}}({{x}^{4}})+c\]
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