Integrate it! Part- II

Let $\displaystyle I = \int_0^{\pi}\frac{4x\sin x}{1+\cos^2x}dx$

We know that,

$\int_a^bf(x) dx = \int_a^bf(a+b-x)dx$

$\Rightarrow I = \int_0^{\pi}\frac{4(\pi-x)\sin x}{1+\cos^2x}dx$

Adding the two $I$ 's,

$2I = 4\pi\int_0^{\pi}\frac{\sin x}{1+\cos^2x}dx$

Substituting $\displaystyle u = \cos x$ ,

$\Rightarrow I = -2\pi\int_1^{-1}\frac{du}{1+u^2}$

$I = -2\pi\left(tan^{-1}u\left.\right|_1^{-1}\right)$

$I = -2\pi(\frac{-\pi}{4}-\frac{\pi}{4}) = -2\pi\left(\frac{-\pi}{2}\right)$

$I = \pi^2 = \boxed{9.869}$

Use this property of integrals. $\int_0^kf(x)\text{ }dx=\int_0^kf(k-x)\text{ }dx$ Let $I=\displaystyle\int_0^\pi\dfrac{4x\sin x}{1+\cos^2x}\text{ }dx.$ Because of the property, $I$ is also equal to $\displaystyle\int_0^\pi\dfrac{4(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}\text{ }dx.$ However, $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x\Rightarrow\cos^2(\pi-x)=\cos^2x,$ we can take those out of the integral. $I=\int_0^\pi\dfrac{4x\sin x}{1+\cos^2x}\text{ }dx=\int_0^\pi\dfrac{4(\pi-x)\sin x}{1+\cos^2x}\text{ }dx$ If you add these integrals together, you will find this. $2I=\int_0^\pi\dfrac{4x\sin x+4(\pi-x)\sin x}{1+\cos^2x}\text{ }dx=\int_0^\pi\dfrac{4\pi\sin x}{1+\cos^2x}\text{ }dx$ $I=2\pi\int_0^\pi\dfrac{\sin x}{1+\cos^2 x}\text{ }dx$ You can solve this integral with a $u\text{-substitution},$ with $u=\cos x$ and $du=-\sin x\text{ }dx.$ $\begin{aligned} 2\pi\int_1^{-1}\dfrac{-du}{1+u^2}&=2\pi\int_{-1}^1\dfrac{du}{1+u^2}\\ &=\left.2\pi\arctan x\right|^1_{-1}\\ &=2\pi\left(\dfrac{\pi}{4}-\dfrac{-\pi}{4}\right)\\ &=2\pi*\dfrac{\pi}{2}\\ &=\pi^2=\boxed{9.869} \end{aligned}$

Here's a solution using integration by parts:

Let $u = 4x$ , $v' = \frac{\sin(x)} {1+\cos^2(x)}$ , so $u' = 4$ and on the substitution $y=\cos (x)$ we have

$v = \int \frac {\sin x} {1+\cos^2 x} dx = -\int \frac{1}{1+y^2} dy = - \arctan(\cos x)$

So our integral $I$ becomes $I = \left[ -4x \arctan( \cos x ) \right] _0 ^\pi + \int_0^ \pi 4 \arctan( \cos x) dx$

On the substitution $y = x - \frac{\pi}{2}$ the integral $\int_0^ \pi 4 \arctan( \cos x) dx$ becomes $\int_{-\pi/2} ^ {\pi/2} 4 \arctan( \sin x) dx$ Which equals 0 since $\arctan( \sin x)$ is odd. Thus:

$I = \left[ -4x \arctan( \cos x ) \right] _0 ^\pi = \pi^2$