Choose the correct answer in the value of $\int\limits_0^\frac{\pi}{2}log\bigg(\frac{\large 4+3\sin x}{\large 4+3\cos x}\bigg)dx$ is

$\begin{array}{1 1}(a)\;2 \\ (b)\;\large\frac{3}{4} \\ (c)\;0 \\ (d)\;-2 \end{array}$

Toolbox:
• (i)$\int \limits_a^bf(x)dx=F(b)-F(a)$
• (ii)$\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
Given $I=\int\limits_0^\frac{\pi}{2}log\bigg[\frac{\large 4+3\sin x}{\large 4+3\cos x}\bigg]dx-----(1)$

Now applying property $\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$

$I=\int\limits_0^\frac{\pi}{2}log\bigg[\frac{\large 4+3\sin(\frac{\pi}{2}-x)}{\large 4+3\cos (\frac{\pi}{2}-x)}\bigg]dx$

But $\sin (\frac{\pi}{2}-x)=\cos x \;and\; \cos(\frac{\pi}{2}-x)=\sin x$

$I=\int\limits_0^\frac{\pi}{2}log\bigg[\frac{\large 4+3\cos x}{\large 4+3\sin x}\bigg]dx-----(2)$

$2I=\int\limits_0^\frac{\pi}{2}\bigg(log\bigg[\frac{\large 4+3\sin x}{\large 4+3\cos x}\bigg]+log\bigg[\frac{\large 4+3\cos x}{\large 4+3\sin x}\bigg]\bigg)dx$

But $log(\frac{a}{b})+log(\frac{c}{d})=log(\frac{a}{b} \times \frac{c}{d}),$similarly

$2I=\int\limits_0^\frac{\pi}{2}log\bigg(\frac{\large 4+3\sin x}{\large 4+3\cos x} \times \frac{\large 4+3\cos x}{\large 4+3\sin x}\bigg)dx$

$\int \limits_0^{\frac{\pi}{2}} log 1 dx$

But $log1=0$

Therfore $2I=0$

$I=0$

Hence the correct answer is C
edited Apr 29, 2016 by meena.p