Evaluate the following integrals:


Evaluate the following integrals:

$\int \frac{1}{1-\cos 2 x} d x$


Let $I=\int \frac{1}{1-\cos 2 x} d x$

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

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

Therefore, we can write the integral as

$I=\int \frac{1}{2 \sin ^{2} x} d x$

$\Rightarrow I=\frac{1}{2} \int \frac{1}{\sin ^{2} x} d x$

$\Rightarrow I=\frac{1}{2} \int \operatorname{cosec}^{2} x d x$

Recall $\int \operatorname{cosec}^{2} x d x=-\cot x+c$

$\Rightarrow \mathrm{I}=\frac{1}{2}(-\cot \mathrm{x})+\mathrm{c}$

$\therefore I=-\frac{1}{2} \cot x+c$

Thus, $\int \frac{1}{1-\cos 2 x} d x=-\frac{1}{2} \cot x+c$

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