Evaluate the following integrals:


Evaluate the following integrals:

$\int \cot ^{-1}\left(\frac{\sin 2 x}{1-\cos 2 x}\right) d x$


Let $I=\int \cot ^{-1}\left(\frac{\sin 2 x}{1-\cos 2 x}\right) d x$

We know $\cos 2 \theta=1-2 \sin ^{2} \theta$

Hence, in the denominator, we can write $1-\cos 2 x=2 \sin ^{2} x$

In the numerator, we have $\sin 2 x=2 \sin x \cos x$

Therefore, we can write the integral as

$I=\int \cot ^{-1}\left(\frac{2 \sin x \cos x}{2 \sin ^{2} x}\right) d x$

$\Rightarrow I=\int \cot ^{-1}\left(\frac{\cos x}{\sin x}\right) d x$

$\Rightarrow I=\int \cot ^{-1}(\cot x) d x$

$\Rightarrow I=\int x d x$

Recall $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$

$\Rightarrow \mathrm{I}=\frac{\mathrm{x}^{1+1}}{1+1}+\mathrm{c}$

$\therefore \mathrm{I}=\frac{\mathrm{x}^{2}}{2}+\mathrm{c}$

Thus, $\int \cot ^{-1}\left(\frac{\sin 2 x}{1-\cos 2 x}\right) d x=\frac{x^{2}}{2}+c$

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