Evaluate $\int\frac{\sin^2 x}{1+\sin^2 x}dx$

Question

$$\frac{\sin^2 x}{1+\sin^2 x}=1-\frac{1}{1+\sin^2 x}$$ so $$\int\frac{\sin^2 x}{1+\sin^2 x}dx=x-\int\frac{dx}{1+\sin^2 x}$$ Let $u=\sin x$, then $du=\cos xdx,\cos x=\sqrt{1-u^2}$, so $$\int\frac{dx}{1+\sin^2 x}=\int\frac{1}{(1+u^2)\cos x}du=\int\frac{1}{(1+u^2)\sqrt{1-u^2}}du$$ I think the answer is related to $\tan^{-1}$, but cannot find a proper substitution.

Answer 1

Euler's Second Substitution, $$u = -\frac{2 t}{1 + t^2}, \qquad du = -\frac{2 (1 - t^2)}{(1 + t^2)^2} \,dt .$$ rationalizes the integral: $$2 \int \frac{1 + t^2}{1 + 6 t^2 + t^4} \,dt .$$ (Just as well, this integral is related to the original integral in $x$ via the tangent half-angle substitution, $x = 2 \arctan t, dx = \frac{2 \,dt}{1 + t^2}$.)

Since the integral in $t$ is rational, the standard method is to decompose the integrand via partial fraction and then integrate each term separately, but the partial fraction decomposition in this case is messy.

We can avoid partial fractions by rewriting the integral as $$2 \int \frac{\frac{1}{t^2} + 1}{\frac{1}{t^2} + 6 + t^2} \,dt = 2 \int \frac{\frac{1}{t^2} + 1}{8 + \left(t - \frac{1}{t}\right)^2} \,dt$$ and substituting $$v = t - \frac{1}{t}, \qquad dv = \left(1 + \frac{1}{t^2}\right) \,dt,$$ which transforms the integral to $$2 \int \frac{dv}{8 + v^2} = \frac{1}{\sqrt 2} \arctan \frac{v}{2 \sqrt 2} + C .$$

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Alternatively, since only even powers of $\sin x$ appear in the integrand, the substitution $$x = \arctan s , \qquad dx = \frac{ds}{1 + s^2} ,$$ also rationalizes the integral:

$$\int \frac{dx}{1 + \sin^2 x} = \int \frac{ds}{2 s^2 + 1} = \frac{1}{\sqrt 2} \arctan (\sqrt 2 s) + C = \boxed{\frac{1}{\sqrt 2} \arctan (\sqrt 2 \tan x) + C}.$$ Notice that the substitution $x = \arctan s$ implicitly assumes that $x \in (-\frac\pi2, \frac\pi2)$, so we should expect that our antiderivative in $x$ is only one on that interval and, by periodicity, on the translates of that interval by integer multiples of $\pi$.

Answer 2

$$\begin{eqnarray} &&\int\frac{\sin^2 x}{1+\sin^2 x}dx=\int\frac{1}{2+\cot^2 x}dx\\ &=&\int\frac{1}{(2+\cot^2 x)(1+\cot^2x)}\csc^2xdx\overset{u=\cot x}=-\int\frac{1}{(2+u^2)(1+u^2)}du\\ &=&-\int\bigg(\frac{1}{1+u^2}-\frac{1}{2+u^2}\bigg)du=-\arctan u+\frac1{\sqrt2}\arctan(\frac u{\sqrt{2}})+C\\ &=&-\arctan(\cot x)+\frac1{\sqrt2}\arctan(\frac{\cot x}{\sqrt{2}})+C \end{eqnarray}$$

Answer 3

Decompose the integrand as follows \begin{align} &\int\frac{\sin^2 x}{1+\sin^2 x}dx\\ =& \int \frac{\sqrt2-1}{\sqrt2}- \frac{\sqrt2-(1+\sin^2x)}{\sqrt2(1+\sin^2 x)}\ dx\\ =& \ \frac{\sqrt2-1}{\sqrt2}x - \frac1{\sqrt2}\tan^{-1}\frac{\sin2x}{(\sqrt2+1)^2-\cos2x}+C \end{align} where the resulting anti-derivative is everywhere continuous.

Answer 4

Observe that from the integrand for your second integral, apply $\cos^2 x+\sin^2 x=1$ to get

$$\frac 1{1+\sin^2 x}=\frac {1}{\cos^2 x+2\sin^2 x}=\frac {\sec^2 x}{1+2\tan^2 x}$$

Let $t=\tan x$ so $\mathrm dt=\sec^2 x\,\mathrm dx$. Thus

$$\begin{align*}\int\left(1-\frac 1{1+\sin^2 x}\right)\,\mathrm dx & =x-\int\frac {\mathrm dt}{1+2t^2}\\ & =x-\frac 1{\sqrt2}\arctan(t\sqrt 2)+C\\ & =x-\frac 1{\sqrt2}\arctan(\sqrt 2\tan x)+C\end{align*}$$

Answer 5

After half-angle substitution $\sin^2x=(1-\cos2x)/2,$ $$\int\frac{\sin^2 x}{1+\sin^2x}dx=\int\left(1-\frac{2}{3-\cos2x}\right)dx$$ Let $t=\tan x$ (Weierstrass Substitution, or Euler's $\tan(\theta/2)$-substitution). Then $\cos2x=\frac{1-t^2}{1+t^2}$ and $dx=\frac{dt}{1+t^2}$ (there is no $2$!). Hence $$\int\frac{\sin^2 x}{1+\sin^2x}dx=x-\int\frac{dt}{2t^2+1}=x-\frac1{\sqrt2}\arctan(\sqrt2 t)+c=x-\frac1{\sqrt2}\arctan(\sqrt2\tan x)+c$$