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Question

Mathematics Question on Integration by Partial Fractions

∫sinx3+4cos2xdx=∫\frac{sinx}{3+4cos^2x}dx=

A

−123tan−1(2cosx3)+C-\frac{1}{2\sqrt 3}tan^{-1}(\frac{2cosx}{\sqrt 3})+C

B

13tan−1(cosx3)+C\frac{1}{\sqrt 3}tan^{-1}(\frac{cosx}{3})+C

C

123tan−1(cosx3)+C\frac{1}{2\sqrt 3}tan^{-1}(\frac{cosx}{3})+C

D

−13tan−1(2cosx3)+C-\frac{1}{\sqrt 3}tan^{-1}(\frac{2cosx}{3})+C

Answer

−123tan−1(2cosx3)+C-\frac{1}{2\sqrt 3}tan^{-1}(\frac{2cosx}{\sqrt 3})+C

Explanation

Solution

The correct answer is Option (A) : −123tan−1(2cosx3)+C-\frac{1}{2\sqrt 3}tan^{-1}(\frac{2cosx}{\sqrt 3})+C