Question

If $ \int_{0}^{\frac{\pi}{3}}\frac{cos x}{3 + 4 sin x}dx = k\, log \left(\frac{3+2\sqrt{3}}{3}\right) $ then $k$ is

• A. $ \frac{1}{2} $
• B. $ \frac{1}{3} $
• C. $ \frac{1}{4} $
• D. $ \frac{1}{8} $

Answer 1

The Correct Option is C

T he correct option is(C): \(\frac{1}{4}\)

We have, 
\(\int_{0}^{\pi /3} \frac{cos\,x}{3+4\,sin\,x} dx\)
\(=k\,log \left(\frac{3+2\sqrt{3}}{3}\right)\ldots\left(i\right)\)
Let \(I=\int_{0}^{\pi/ 3} \frac{cos\,x}{3+4\,sin\,x}dx\)
Put \(3 + 4 sin \,x = t\)
\(\Rightarrow 0+4\,cos\,x\,dx =dt\)
Upper limit, \(x=\frac{\pi}{3}, t=3+4\,sin \frac{\pi}{3}\)
\(=3+4\times\frac{\sqrt{3}}{2}=3+2\sqrt{3}\)
and lower limit \(x=0\), 
\(t=3+4\,sin \, 0=3\)
\(\therefore I=\int_{3}^{3+2\sqrt{3}} \frac{dt}{4t}=\frac{1}{4} \left[log\,t\right]_{3}^{3+2\sqrt{3}}\)
\(=\frac{1}{4}\left[log\left(3+2\sqrt{3}\right)-log\,3\right]\)
\(\Rightarrow I=\frac{1}{4} log \left(\frac{3+2\sqrt{3}}{3}\right) \ldots\left(ii\right)\)
\(\therefore\) From Eqs. \(\left(i\right)\) and \(\left(ii\right)\), we get 
\(k=\frac{1}{4}\)