find $I=\int\frac{\sin^2(x)dx}{\sin(x)+2\cos(x)}$

Question

Find the integral $$I=\int\dfrac{\sin^2(x)dx}{\sin(x)+2\cos(x)}$$

Putting $t=\tan\dfrac{x}{2}\Rightarrow x=2\arctan(t), dx=\dfrac{2dt}{1+t^2},\sin(x)=\dfrac{2t}{1+t^2},\cos(x)=\dfrac{1-t^2}{1+t^2}$ gives $$\begin{align*} I &= \int\dfrac{\sin^2(x)dx}{\sin(x)+2\cos(x)} \\ &= \int\dfrac{\frac{4t^2}{(1+t^2)^2}\frac{2dt}{1+t^2}}{\frac{2t}{1+t^2}+\frac{2(1-t^2)}{1+t^2}}=\int\dfrac{\frac{8t^2dt}{(1+t^2)^3}}{\frac{2t+2-2t^2}{1+t^2}}=\int\dfrac{\frac{4t^2dt}{(1+t^2)^2}}{-t^2+t+1}=\int\dfrac{4t^2}{(1+t^2)^2(-t^2+t+1)}dt, \end{align*} $$ but this doesn't seem fun to integrate even thought it's a rational function. Is there a better approach (maybe a better substitution)?

Answer 1

You can decompose the integrand from the start as

$$\begin{align*} \frac{\sin^2x}{\sin x+2\cos x} &= \frac{\sin^2x(\sin x-2\cos x)}{\sin^2x-4\cos^2x} \\ &= \frac{\sin^2x(\sin x-2\cos x)}{5\sin^2x-4} \\ &= \frac15 (\sin x-2\cos x) - \frac85 \cdot \frac{\cos x}{5\sin^2x-4} + \frac45 \cdot \frac{\sin x}{5\sin^2x-4} \end{align*}$$

Integrating the first two terms is easy. For the last term, employing the half-angle tangent sub leaves you with

$$\int \frac{\sin x}{5\sin^2x-4} \, dx \stackrel{x=2\arctan t}= - \int \frac{t}{t^4-3t^2+1} \, dt$$

and integrating becomes trivial upon completing the square in the denominator.

Alternatively, as OP pointed out in comments below, we have

$$\frac{\sin x}{5\sin^2x-4} = \frac{\sin x}{3-5\cos^2x}$$

so using the tangent substitution for this third term is in fact overkill.

Answer 2

To solve the integral $I = \int \frac{\sin^2(x)}{\sin(x) + 2\cos(x)} \,dx$, we can start by simplifying the integrand.

First, notice that the denominator has both $ \sin(x) $ and $ \cos(x) $. Let's express everything in terms of $ \sin(x) $ to make it easier:

$ I = \int \frac{\sin^2(x)}{\sin(x) + 2\cos(x)} \,dx $

Now, use the identity $ \cos(x) = \sqrt{1 - \sin^2(x)} $ to replace $ \cos(x) $:

$ I = \int \frac{\sin^2(x)}{\sin(x) + 2\sqrt{1 - \sin^2(x)}} \,dx $

Next, let's substitute $ u = \sin(x) $ and $ du = \cos(x) \,dx $:

$ I = \int \frac{u^2}{u + 2\sqrt{1 - u^2}} \,du $

Now, factor the denominator:

$ I = \int \frac{u^2}{u + 2\sqrt{1 - u^2}} \,du = \int \frac{u^2}{u + 2\sqrt{1 - u^2}} \cdot \frac{u - 2\sqrt{1 - u^2}}{u - 2\sqrt{1 - u^2}} \,du $

Combine the terms in the numerator:

$ I = \int \frac{u^3 - 2u^2\sqrt{1 - u^2}}{u^2 - 4(1 - u^2)} \,du $

Simplify the denominator:

$ I = \int \frac{u^3 - 2u^2\sqrt{1 - u^2}}{5u^2 - 4} \,du $

Now, perform partial fraction decomposition to break this into simpler fractions:

$ I = \int \left(\frac{1}{5} - \frac{2u}{5(u^2 - 4)} - \frac{2\sqrt{1 - u^2}}{5(u^2 - 4)}\right) \,du $

Now, you can integrate each term separately:

$ I = \frac{1}{5}u - \frac{1}{5}\ln|u + 2| - \frac{1}{5}\arcsin\left(\frac{u}{2}\right) + C $

Finally, substitute back $ u = \sin(x) $:

$ I = \frac{1}{5}\sin(x) - \frac{1}{5}\ln|\sin(x) + 2| - \frac{1}{5}\arcsin\left(\frac{\sin(x)}{2}\right) + C $

where $ C $ is the constant of integration.