- $\pi$
- $\frac{\pi}{2}$
- $0$
- $2\pi$
Solution:
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x }\log\tan\text{x dx}\ ....(\text{i})$
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin(\pi-2\text{x})\log\tan\big(\frac{\pi}{2}-\text{x}\big)\text{dx}$
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\log\cot\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}+\log\cot\text{x}\big)\text{dx}$
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}\cot\text{x}\big)\text{dx}$
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin\text{x}(\log1)\text{dx}$
$\text{I}=0$
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Then, the number of functions $g:[-1,1] \rightarrow[0,1]$ such that $(g \circ f)(x)=x$ for all $x \in[0,1]$ is