Question
Choose the correct answer in Exercise: The value of $\int^{\frac{\pi}{2}}\limits_{0}\log\bigg(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\bigg)\text{dx}\ $is
-
2
- $\frac{3}{4}$
- 0
- -2
2
$\text{Let}\ \text{I}=\int^{\frac{\pi}{2}}\limits_{0}\log\bigg(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\bigg)\text{dx}\ \ \ ....(1)$
$\Rightarrow\ \ \ \text{I}=\int^{\frac{\pi}{2}}\limits_{0}\log\begin{bmatrix} \frac{4+3\sin\bigg(\frac{\pi}{2}-\text{x}\bigg)}{4+3\cos\bigg(\frac{\pi}{2}-\text{x}\bigg)}\text{dx}\\ \end{bmatrix}\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}\log\bigg[\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\bigg]\text{dx}\ \ \ ....(2)$
Adding eq. (i) and (ii),
$2\text{I}=\int^{\frac{\pi}{2}}\limits_{0}\bigg[\log\bigg(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\bigg)+\log\bigg(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\bigg)\bigg]\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}\bigg[\log\bigg(\frac{4+3\sin\text{x}}{4+3\sin\text{x}}\bigg)\bigg(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\bigg)\bigg]\text{dx}$
$\Rightarrow\ \ \ 2\text{I}=\int^{\frac{\pi}{2}}\limits_{0}(\log1)\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}0\ \text{dx}=0\ \ \ \ \ \Rightarrow\text{I}=0$
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