# Determine the following indefinite integral. \int \frac{dx}{1-\sin^{2}x}

Determine the following indefinite integral.
$$\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{1}-{{\sin}^{{{2}}}{x}}}}}$$

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Step 1
Given integral:
$$\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{1}-{{\sin}^{{{2}}}{x}}}}}$$
Step 2
Now,
Use the following identity: $$\displaystyle{1}-{{\sin}^{{{2}}}{\left({x}\right)}}={{\cos}^{{{2}}}{\left({x}\right)}}$$
$$\displaystyle\int{\frac{{{1}}}{{{1}-{{\sin}^{{{2}}}{\left({x}\right)}}}}}{\left.{d}{x}\right.}=\int{\frac{{{1}}}{{{{\cos}^{{{2}}}{\left({x}\right)}}}}}{\left.{d}{x}\right.}$$
Use the following identity: $$\displaystyle{\frac{{{1}}}{{{\cos{{\left({x}\right)}}}}}}={\sec{{\left({x}\right)}}}$$
$$\displaystyle=\int{{\sec}^{{{2}}}{\left({x}\right)}}{\left.{d}{x}\right.}$$
Use the common integral: $$\displaystyle\int{{\sec}^{{{2}}}{\left({x}\right)}}{\left.{d}{x}\right.}={\tan{{\left({x}\right)}}}$$
$$\displaystyle={\tan{{\left({x}\right)}}}$$
Add a constant to the solution:
$$\displaystyle={\tan{{\left({x}\right)}}}+{C}$$