# Asking for help on a question about integral: $\int_0^\pi \frac{x\sin x}{3+\cos^2 x}\mathrm{d}x$

Given the following intergral, and the fact that $$a$$ and $$c$$ are prime numbers,

$$\int_0^\pi \frac{x\sin x}{3+\cos^2 x}\mathrm{d}x= \frac{π^a}{b\sqrt c}$$

Evaluate $$a+b+c$$

I've tried to solve the intergral by using intergration by part where I let $$u$$ be $$x$$ and $$dv$$ be $$\frac{\sin x}{3+\cos^2 x}$$. Which then gave me

$$\int_0^\pi \frac{x\sin x}{3+\cos^2 x}\mathrm{d}x = \frac{-x}{\sqrt 3}\arctan{\left(\frac{\cos x}{\sqrt 3}\right)}\Big|_0^π + \int_0^\pi \frac{1}{\sqrt 3}\arctan{\left(\frac{\cos x}{\sqrt 3}\right)}\mathrm{d}x$$

But I'm pretty much out of luck at this point since I don't know how to solve $$\int_0^π v\mathrm{d}u$$ in this case and I believe there is a better way to do this.

• As much as I remember, this question was featured in Brilliant integration challenges, right? Oct 10, 2021 at 2:05

Hint...there is a useful trick you could try using here:

$$I=\int_0^{\pi}xf(\sin x)dx=\pi\int_0^{\pi}f(\sin x)dx-I$$

which is readily seen if you substitute $$u=\pi-x$$ into the first integral...

• $+1$ Literally I already knew that but realized now, Lol!
– user960916
Oct 9, 2021 at 18:06

So I eventually figured out the answer to $$I_1 = \int_0^\pi \frac{x\sin x}{3+\cos^2 x}\mathrm{d}x$$ By using the following property: $$\int_a^b f(x)\mathrm{d}x = \int_a^b f(a+b-x)\mathrm{d}x$$ we can proof that $$I_1 = \frac{π}{2}\int_0^π f(\sin x)\mathrm{d}x = \frac{π}{2}I_2$$ where $$I_2 = \int_0^π \frac{\sin x}{3+\cos^2 x}\mathrm{d}x$$. To solve for $$I_2$$ we substitute $$u$$ for $$\cos x$$ which gave us $$I_2=\frac{1}{\sqrt 3}\arctan \left(\frac{u}{\sqrt 3}\right) \big|_{-1}^1=\frac{π\sqrt 3}{9}$$ Therefore $$I_1=\frac{π}{2}\frac{π\sqrt 3}{9}=\frac{π^2}{6\sqrt 3}$$ And $$a+b+c = 11$$

Your first attempt with integration by parts also works out nicely, just follow up substituting $$x\mapsto\dfrac\pi2-x$$ to see the remaining integral vanishes:

\begin{align*} & \int_0^\pi \frac{x \sin x}{3 + \cos^2x} \, dx \\ &= \frac{\pi^2}{6\sqrt3} + \frac1{\sqrt3} \int_0^\pi \arctan \left(\frac{\cos x}{\sqrt3}\right) \, dx \\ &= \boxed{\frac{\pi^2}{6\sqrt3}} + \frac1{\sqrt3} \underbrace{\int_{-\tfrac\pi2}^{\tfrac\pi2} \arctan\left(\frac{\sin x}{\sqrt3}\right) \, dx}_{=0} \end{align*}