# Find the integration by using substitution method. \int \frac{\cot x}{\sin x}dx

## Question:

Find the integration by using substitution method.

{eq}\displaystyle \int \frac{\cot x}{\sin x}dx {/eq}

## Substitution by Trig-properties:

Firstly, we'll change the given integral expression in terms of sine and cosine function using the general formula of cotangent function.

• {eq}\displaystyle \cot \theta=\frac{\cos \theta}{\sin \theta} {/eq}

After that, substitute the trig-function of the denominator of the obtained expression as {eq}u {/eq} and use the common derivative formula.

• {eq}\displaystyle \frac{\mathrm{d} (\sin x)}{\mathrm{d} x}=\cos x {/eq}

The given integral expression is:

{eq}I=\displaystyle \int \frac{\cot x}{\sin x}\ dx {/eq}

Using the formula of cotangent function in terms of sine and cosine function in the above expression, we get:

{eq}\begin{align*} \displaystyle I&=\int \frac{\frac{\cos x}{\sin x}}{\sin x}\ dx\\ &=\int \frac{\cos x}{\sin x\cdot \sin x}\ dx\\ &=\int \frac{\cos x}{\sin ^2x}\ dx.............(1) \end{align*} {/eq}

For substitution, we'll take {eq}u= \sin x {/eq}

Differentiating both sides of the above expression with their respective variables, we get:

{eq}\ du= \cos x\ dx {/eq}

Using the above values in equation (1), we get:

{eq}\begin{align*} \displaystyle I&=\int \frac{1}{u^2}\ du\\ &=\int u^{-2}\ du&\because \frac{1}{t^n}=t^{-n}\\ &=\frac{u^{-2+1}}{-2+1}+C\\ &=\frac{u^{-1}}{-1}+C\\ &=-\frac{1}{u}+C\\ \end{align*} {/eq}

Where,

• {eq}C {/eq} is the constant of integration.

Back substitute the value of {eq}u {/eq} in the above expression and simplify it.

{eq}\displaystyle I=-\frac{1}{\sin x}+C {/eq} 