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Question

Mathematics Question on Definite Integral

∫x3sin⁡(tan⁡−1(x4))1+x8dx\int\frac{x^3\sin(\tan^{-1}(x^4))}{1+x^8}dx is equal to

A

−cos⁡(tan⁡−1(x4))4+C\frac{-\cos(\tan^{-1}(x^4))}{4}+C

B

cos⁡(tan⁡−1(x4))4+C\frac{\cos(\tan^{-1}(x^4))}{4}+C

C

−cos⁡(tan⁡−1(x3))3+C\frac{-\cos(\tan^{-1}(x^3))}{3}+C

D

−sin⁡(tan⁡−1(x4))4+C\frac{-\sin(\tan^{-1}(x^4))}{4}+C

Answer

−cos⁡(tan⁡−1(x4))4+C\frac{-\cos(\tan^{-1}(x^4))}{4}+C

Explanation

Solution

The correct answer is (A) : −cos⁡(tan⁡−1(x4))4+C\frac{-\cos(\tan^{-1}(x^4))}{4}+C.