Evaluate the following integrals:


Evaluate the following integrals:

$\int \frac{\operatorname{cosec} x}{\operatorname{cosec} x-\cot x} d x$


Let $I=\int \frac{\operatorname{cosec} x}{\operatorname{cosec} x-\cot x} \mathrm{dx}$

On multiplying and dividing $(\operatorname{cosec} x+\cot x)$, we can write the integral as

$I=\int \frac{\operatorname{cosec} x}{\operatorname{cosec} x-\cot x}\left(\frac{\operatorname{cosec} x+\cot x}{\operatorname{cosec} x+\cot x}\right) d x$

$\Rightarrow I=\int \frac{\operatorname{cosec} x(\operatorname{cosec} x+\cot x)}{(\operatorname{cosec} x-\cot x)(\operatorname{cosec} x+\cot x)} d x$

$\Rightarrow I=\int \frac{\operatorname{cosec}^{2} x+\operatorname{cosec} x \cot x}{\operatorname{cosec}^{2} x-\cot ^{2} x} d x$

$\Rightarrow I=\int\left(\operatorname{cosec}^{2} x+\operatorname{cosec} x \cot x\right) d x\left[\because \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\right]$

$\Rightarrow I=\int \operatorname{cosec}^{2} x d x+\int \operatorname{cosec} x \cot x d x$

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

We also have $\int \operatorname{cosecx} \cot x d x=-\operatorname{cosec} x+c$

$\therefore I=-\cot x-\operatorname{cosec} x+c$

Thus, $\int \frac{\operatorname{cosec} x}{\operatorname{cosec} x-\cot x} d x=-\cot x-\operatorname{cosec} x+c$

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