Question

Integrate $\int \frac{1}{1+cotx}dx$

Answer 1

Solve the given integral

Given integral,

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

We know that ,

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

Now,

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

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

Where , $$\begin{gathered} u=\sin x+\cos x\mathrm{and} \\ du=(\cos x-\sin x)dx \end{gathered}$$

Now,

$$\begin{gathered} =\frac{x}{2}-\frac{1}{2}\ln \left|u\right|+C \\ =\frac{x}{2}-\frac{1}{2}\ln |\sin x+\cos x|+C \end{gathered}$$

Hence , Integral $\int \frac{1}{1+cotx}dx$ is $\frac{x}{2}-\frac{1}{2}\ln |\sin x+\cos x|+C$.