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$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\operatorname{cosec} x \cdot \cot x}{1+\operatorname{cosec}^2 x} d x=$
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Verified Answer
The correct answer is:
$\tan ^{-1}\left(\frac{1}{3}\right)$
Let $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\operatorname{cosec} x \cdot \cot x}{1+\operatorname{cosec}^2 x} d x$
Put $\operatorname{cosec} x=t \Rightarrow \operatorname{cosec} x \cot x=-d t$. When $x=\frac{\pi}{6}, t=2$ and when $x=\frac{\pi}{2}, t=1$
$$
\begin{aligned}
& \mathrm{I}=\int_2^1(-1) \frac{\mathrm{dt}}{1+\mathrm{t}^2} \\
& =\int_1^2 \frac{\mathrm{dt}}{1+\mathrm{t}^2}=\left[\tan ^{-1} \mathrm{t}_1^2=\tan ^{-1}(2)-\tan ^{-1}(1)=\tan ^{-1}\left[\frac{2-1}{1+(2)(1)}\right]=\tan ^{-1}\left(\frac{1}{3}\right)\right.
\end{aligned}
$$
Put $\operatorname{cosec} x=t \Rightarrow \operatorname{cosec} x \cot x=-d t$. When $x=\frac{\pi}{6}, t=2$ and when $x=\frac{\pi}{2}, t=1$
$$
\begin{aligned}
& \mathrm{I}=\int_2^1(-1) \frac{\mathrm{dt}}{1+\mathrm{t}^2} \\
& =\int_1^2 \frac{\mathrm{dt}}{1+\mathrm{t}^2}=\left[\tan ^{-1} \mathrm{t}_1^2=\tan ^{-1}(2)-\tan ^{-1}(1)=\tan ^{-1}\left[\frac{2-1}{1+(2)(1)}\right]=\tan ^{-1}\left(\frac{1}{3}\right)\right.
\end{aligned}
$$
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