Question

$∫\frac{sinx}{3+4cos^2x}dx=$

• A. $-\frac{1}{2\sqrt 3}tan^{-1}(\frac{2cosx}{\sqrt 3})+C$
• B. $\frac{1}{\sqrt 3}tan^{-1}(\frac{cosx}{3})+C$
• C. $\frac{1}{2\sqrt 3}tan^{-1}(\frac{cosx}{3})+C$
• D. $-\frac{1}{\sqrt 3}tan^{-1}(\frac{2cosx}{3})+C$

Answer 1

The Correct Option is A

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