Question:
\(∫\frac{sinx}{3+4cos^2x}dx=\)
\(∫\frac{sinx}{3+4cos^2x}dx=\)
Updated On: Apr 26, 2024
- \(-\frac{1}{2\sqrt 3}tan^{-1}(\frac{2cosx}{\sqrt 3})+C\)
- \(\frac{1}{\sqrt 3}tan^{-1}(\frac{cosx}{3})+C\)
- \(\frac{1}{2\sqrt 3}tan^{-1}(\frac{cosx}{3})+C\)
- \(-\frac{1}{\sqrt 3}tan^{-1}(\frac{2cosx}{3})+C\)
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The Correct Option is A
Solution and Explanation
The correct answer is Option (A) : \(-\frac{1}{2\sqrt 3}tan^{-1}(\frac{2cosx}{\sqrt 3})+C\)
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