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Question

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

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Solution

## Step: 1 Simplify the given integralGiven, $\int \frac{1}{1+cotx}dx$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,$cotx=\frac{\mathrm{cos}x}{\mathrm{sin}x}$Now solve, $\int \frac{1}{1+cotx}dx$$=\int \left(\frac{1}{1+\frac{\mathrm{cos}x}{\mathrm{sin}x}}\right)dx\phantom{\rule{0ex}{0ex}}=\int \left(\frac{\mathrm{sin}x}{\mathrm{sin}x+\mathrm{cos}x}\right)dx\phantom{\rule{0ex}{0ex}}=\int \frac{1}{2}\left(\frac{2\mathrm{sin}x}{\mathrm{sin}x+\mathrm{cos}x}\right)dx\phantom{\rule{0ex}{0ex}}=\int \frac{1}{2}\left\{\frac{\left(\mathrm{sin}x+\mathrm{cos}x\right)\left(\mathrm{sin}x-\mathrm{cos}x\right)}{\mathrm{sin}x+\mathrm{cos}x}\right\}dx\phantom{\rule{0ex}{0ex}}=\int \frac{1}{2}dx+\int \frac{1}{2}\left\{\frac{\mathrm{sin}x-\mathrm{cos}x}{\mathrm{sin}x+\mathrm{cos}x}\right\}dx\phantom{\rule{0ex}{0ex}}$Step: 2 Using the Substitution method to find the value of integralLet assume, $\mathrm{sin}x+\mathrm{cos}x=t$On differentiating we get,$\left(\mathrm{cos}x-\mathrm{sin}x\right)dx=dt$$\left(\mathrm{sin}x-\mathrm{cos}x\right)dx=-dt$Now,$=\int \frac{1}{2}dx+\frac{1}{2}\int \frac{-dt}{t}\phantom{\rule{0ex}{0ex}}=\frac{x}{2}-\frac{1}{2}\mathrm{log}t+C$On substituting again, we get,$=\frac{x}{2}-\frac{1}{2}\mathrm{log}\left(\mathrm{sin}x+\mathrm{cos}x\right)+C$Hence, the Integral of $\int \frac{1}{1+cotx}dx$ is $\frac{x}{2}-\frac{1}{2}\mathrm{log}\left(\mathrm{sin}x+\mathrm{cos}x\right)+C$, where $C$ is an arbitrary constant.

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