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Evaluate the integral of (sin2 x) / (1 + sin2 x)

by RoRi

Compute the following integral.

    \[ \int \frac{\sin^2 x}{1+\sin^2 x} \, dx. \]

First, we rewrite the integrand as

    \[ \frac{\sin^2 x}{1+\sin^2 x} = \frac{1+\sin^2 x - 1}{1+\sin^2 x} = 1 - \frac{1}{1+\sin^2 x}. \]

Then we have

    \begin{align*} \int \frac{\sin^2 x}{1+\sin^2 x} \, dx &= \int dx - \int \frac{1}{1+\sin^2 x} \, dx \\[9pt] &= x - \int \frac{\csc^2 x}{\csc^2 x + 1} \, dx \\[9pt] &= x - \int \frac{\csc^2 x}{2 + \cot^2 x} \, dx. \end{align*}

Then we make the substitution u = \cot x which implies du = -\csc^2 x \, dx. Therefore,

    \begin{align*} \int \frac{\sin^2 x }{1+\sin^2 x} \, dx &= x - \int \frac{\csc^2 x}{2+\cot^2 x} \, dx \\[9pt] &= x + \int \frac{du}{2+u^2} \\[9pt] &= x + \frac{1}{2} \int \frac{du}{1+ \left( \frac{u}{\sqrt{2}} \right)^2} \\[9pt] &= x + \frac{1}{\sqrt{2}} \arctan \left( \frac{u}{\sqrt{2}} \right) + C \\[9pt] &= x + \frac{1}{\sqrt{2}} \arctan \frac{\cot x}{\sqrt{2}} + C \\[9pt] &= x - \frac{1}{\sqrt{2}} \arctan (\sqrt{2} \tan x) + C. \end{align*}

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