Question

The value of $k$ if ${\int }_0^{\pi /3}\frac{\cos x}{3+4\sin x}dx=k.\log \left(\frac{3+2\sqrt{3}}{3}\right)$. is equal to

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

Answer 1

Answer: (C)

$\frac{1}{4}$

${\int }_0^{\frac{\pi }{3}}\frac{cosx}{3+4sinx}dxLet3+4sinx=t⟶\left(1\right)\Rightarrow 0+4cosx=\frac{dt}{dx}(diffrentiatew.r.tx)\Rightarrow cosxdx=\frac{dt}{4}forx=0,3+4sin0=t\Rightarrow t=3and,forx=\frac{\pi }{3},3+4sin\frac{\pi }{3}=t\Rightarrow t=3+2\sqrt{3}\therefore {\int }_0^{\frac{\pi }{3}}\frac{cosx}{3+4sinx}dx={\int }_3^{3+2\sqrt{3}}\frac{1}{t}.\frac{dt}{4}=\frac{1}{4}{\left[logt\right]}_3^{3+2\sqrt{3}}=\frac{1}{4}\left[log\right(3+2\sqrt{3})-log3]=\frac{1}{4}log\left(\frac{3+2\sqrt{3}}{3}\right)⟶\left(2\right)\because {\int }_0^{\frac{\pi }{3}}\frac{cosx}{3+4sinx}dx=klog\left(\frac{3+2\sqrt{3}}{3}\right)\Rightarrow \frac{1}{4}log\left(\frac{3+2\sqrt{3}}{3}\right)=klog\left(\frac{3+2\sqrt{3}}{3}\right)comparingbothsideweget,\therefore k=\frac{1}{4}$