Question

Evaluate the Integral \(\int {cosec\,x\,dx} \)

Short answer

Hence we get\(\int {cosec\,x\,dx} = log\left| {cosec\,x - cot\,x} \right| + c\)

Step-by-step solution

Using the substitution method.

Use substitution method and \(cosec\,x - cot\,x = t\)

\(I = \int {\frac{{cosecx\left( {cosecx - cotx} \right)}}{{\left( {cosecx - cotx} \right)}}} dx\)

Put \(cosec\,x - cot\,x = t\)\(\)

Differentiating both sides with respect to x

\(\begin{aligned}{l}\left( { - cosec\,x\,cot\,x + cose{c^2}x} \right)dx &= dt\\dx &= \frac{{dt}}{{ - cosecx\left( {cotx - cosecx} \right)}}\end{aligned}\)

Put the value of \(dx\) in integral I

Therefore, \(I = \int {\frac{{cosecx\left( {cosecx - cotx} \right)}}{{\left( {cosecx - cotx} \right)}}} \times \frac{{dt}}{{ - cosecx\left( {cotx - cosecx} \right)}}\)

Substitute the values and simplify.

\(I = \int {\frac{{cosecx\left( {cosecx - cotx} \right)}}{{\left( {cosecx - cotx} \right)}}} \times \frac{{dt}}{{ - cosecx\left( {cotx - cosecx} \right)}}\)

\(I = \frac{{dt}}{{\left( {cotx - cosecx} \right)}}\)

we know \(cosec\,x - cot\,x = t\)

Substitute the value\(I = \int {\frac{{dt}}{t}} \)

Integration of \(\frac{1}{t} = log\left| t \right| + c\)

Therefore, \(I = log\left| t \right| + c\)

We know, \(t = cosec\,x - cot\,x\)

So, \(I = \log \left| {cosec\,x - cot\,x} \right| + c\)

Hence, \(I = \int {cosec\,x\,dx = } log\left| {cosec\,x - cot\,x} \right| + c\)