Question

Integrate $\int \frac{1}{1+cotx}dx$ using substitution method.

Answer 1

Step: 1 Simplify the given integral

Given, $\int \frac{1}{1+cotx}dx$

We will solve the equation by substitution method. In this method of integration by substitution, any given integral is transformed into a simple form of integral by substituting the independent variable by others.

We know that,

$cotx=\frac{\cos x}{\sin x}$

Now solve, $\int \frac{1}{1+cotx}dx$

$$\begin{gathered} =\int \left(\frac{1}{1+{\displaystyle \frac{\cos x}{\sin x}}}\right)dx \\ =\int \left(\frac{\sin x}{\sin x+\cos x}\right)dx \\ =\int \frac{1}{2}\left(\frac{2\sin x}{\sin x+\cos x}\right)dx \\ =\int \frac{1}{2}\left\{\frac{\left(\sin x+\cos x\right)\left(\sin x-\cos x\right)}{\sin x+\cos x}\right\}dx \\ =\int \frac{1}{2}dx+\int \frac{1}{2}\left\{\frac{\sin x-\cos x}{\sin x+\cos x}\right\}dx \end{gathered}$$

Step: 2 Using the Substitution method to find the value of integral

Let assume, $\sin x+\cos x=t$

On differentiating we get,

$(\cos x-\sin x)dx=dt$

$(\sin x-\cos x)dx=-dt$

Now,

$$\begin{gathered} =\int \frac{1}{2}dx+\frac{1}{2}\int \frac{-dt}{t} \\ =\frac{x}{2}-\frac{1}{2}\log t+C \end{gathered}$$

On substituting again, we get,

$=\frac{x}{2}-\frac{1}{2}\log \left(\sin x+\cos x\right)+C$

Hence, the Integral of $\int \frac{1}{1+cotx}dx$ is $\frac{x}{2}-\frac{1}{2}\log \left(\sin x+\cos x\right)+C$, where $C$ is an arbitrary constant.