Integral from yesterdays test: $\int_0^\pi {{1+\sin^2x}\over{6-\cos^2x+\left|\cos x\right|}} \sin x \cdot x \, dx$

Let me write $$f(x)={{1+\sin^2x}\over{6-\cos^2x+\left|\cos x\right|}} \sin x$$ and $$I=\int_0^\pi f(x) x\,dx.$$ Since $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$, we have $f(\pi-x)=f(x)$. So, making the substitution $u=\pi-x$ in the integral, we have $$I=\int_{\pi}^0f(u)(\pi-u)(-du)=\pi\int_0^\pi f(u)\,du-\int_0^\pi f(u)u\,du=\pi\int_0^\pi f(u)\,du-I$$ and so $$2I=\pi\int_0^\pi f(u)\,du.$$ The identity $f(\pi-x)=f(x)$ furthermore gives that $$\int_0^\pi f(x)\,dx=2\int_0^{\pi/2}f(x)\,dx.$$ So, it suffices to compute the integral $$J=\int_0^{\pi/2}f(x)\,dx$$ and then we will have $I=\pi J$.

To compute $J$, first observe that $\cos x$ is always nonnegative for $x\in[0,\pi/2]$ so we may ignore the absolute value in $f(x)$. We now make the substitution $u=\cos x$ to get $$J=\int_1^0\frac{2-u^2}{6-u^2+u}(-du)=\int_0^1\frac{2-u^2}{-(u-3)(u+2)}\,du$$ (here we use the fact that $\sin^2 x=1-\cos^2 x$ to write the numerator in terms of $u$). Now we use partial fractions to write the integrand as $$1+\frac{7/5}{u-3}-\frac{2/5}{u+2}$$ which we can now integrate straightforwardly to find that $$J=1+\frac{7}{5}(\log2-\log 3)-\frac{2}{5}(\log 3-\log 2)=1+\frac{9}{5}\log\frac{2}{3}.$$

We conclude that $$I=\pi\left(1+\frac{9}{5}\log\frac{2}{3}\right).$$

(For confirmation that I have managed to not make any small mistake, WolframAlpha computes both the integral and my answer to be approximately $0.848741$.)

I was about to say there is a certain symmetry that implies that the integral from $0$ to $\pi/2$ cancels out the integral from $\pi/2$ to $\pi,$ since the function is symmetric about the point $(\pi/2,0).$ But that is not true, because of the multiplication by $x,$ so just skip this paragraph unless you want to see if you can figure out a subtler version of that argument.

However: $$ {{1+\sin^2x}\over{6-\cos^2x+\left|\cos x\right|}} \sin x \, dx = \frac{2-\cos^2 x}{6-\cos^2 x + \left|\cos x\right|} (\sin x\,dx) = \frac{2-w^2}{6-w^2+|w|}\,dw $$ On the interval from $0$ to $\pi/2$ you have $\cos x\ge0$ so you can drop the absolute value sign. Then you have $6-w^2+w=-(w-3)(w+2).$ And on the interval from $\pi/2$ to $\pi$ you have $6-w^2-w = -(w+3)(w-2).$

So first use partial fractions and write $du = \dfrac{2-w^2}{6-w^2+w}\,dw$ or else the version with $6-w^2-2,$ and integrate to find out what $u$ is, and then you have $$ \int x \left( \frac{2-w^2}{6-w^2+w} \, dw \right) = \underbrace{\int x\, du = xu - \int u\, dx}_\text{integration by parts} $$ and work separately on the interval from $\pi/2$ to $\pi,$ in a similar way.