Definite integral $\int _{-1}^1\:\frac{\arccos\left(x\right)}{1+x^2}dx$

Question

Do you have any idea about how should I approach this integral? I tried various substitutions and ended up nowhere. $$\int _{-1}^1\:\frac{\arccos\left(x\right)}{1+x^2}dx$$

Answer 1

Substitute $x\mapsto-x$, then average the two integrals: $$ \begin{align} \int_{-1}^1\frac{\arccos(x)}{1+x^2}\,\mathrm{d}x &=\int_{-1}^1\frac{\pi-\arccos(x)}{1+x^2}\,\mathrm{d}x\\ &=\frac\pi2\int_{-1}^1\frac1{1+x^2}\,\mathrm{d}x\\ &=\frac{\pi^2}4 \end{align} $$

Answer 2

First of all make the substitution $$x=\cos t\implies dx=-\sin t\cdot dt$$ And your integral becomes: $$\int_{-1}^{1}\frac{\arccos(x)}{1+x^2}dx=\int_{\pi}^{0}\frac{\arccos(\cos t)}{1+(\cos t)^2}(-\sin t)dt=\\=-\int_{\pi}^{0}\frac{t\cdot\sin t}{1+(\cos t)^2}dt=\int_{0}^{\pi}\frac{t\cdot\sin t}{1+(\cos t)^2}dt=\color{red}{\frac{\pi^2}{4}}$$

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Now I will show you how to compute the last step.

In order to compute $$I=\int_{0}^{\pi}\frac{t\cdot\sin t}{1+(\cos t)^2}dt$$ we have first to make the substitution $y=\pi -x\implies dx=-dy$ Then $$I=\int_{\pi}^{0}\frac{(\pi-y)\sin (\pi-y)}{1+(\cos (\pi -y))^2}(-dy)=\\=\int_{\pi}^{0}\frac{(\pi-y)[\sin(\pi)\cos(y)-\cos(\pi)\sin(y)]}{1+[\cos(\pi) \cos(y)-\sin(\pi)\sin(y)]^2}(-dy)=\int_{0}^{\pi}\frac{(\pi-y)\cdot\sin y}{1+(\cos y)^2}dy$$ Now split it into two integrals and we see that $$I=\int_{0}^{\pi}\frac{(\pi-y)\cdot\sin y}{1+(\cos y)^2}dy=\\=\pi\int_{0}^{\pi}\frac{\sin y}{1+(\cos y)^2}dy-\int_{0}^{\pi}\frac{y\sin y}{1+(\cos y)^2}dy$$ That is $$I=\pi\int_{0}^{\pi}\frac{\sin y}{1+(\cos y)^2}dy-I$$ or, in other words, $$I=\frac{\pi}2\int_{0}^{\pi}\frac{\sin y}{1+(\cos y)^2}dy$$ Now we havo to make a second substitution: $u=\cos y\implies dy=-\frac{du}{\sin y}$

$$I=\frac{\pi}2\int_{1}^{-1}\frac{\sin y}{1+u^2}\left(-\frac{du}{\sin y}\right)=-\frac{\pi}2\int_{1}^{-1}\frac {du}{1+u^2}=\frac{\pi}2\left[\arctan(u)\right]_{-1}^{1}=\color{red}{\frac{\pi^2}{4}}$$

Answer 3

Hint: using integration by part you will get :

$$\int _{-1}^1\:\frac{\arccos\left(x\right)}{1+x^2}dx= [_{-1}^{1} \arccos x\arctan x +\int_{-1}^{1} \frac{\arctan x}{\sqrt[] {1-x²} }=\frac{\pi²}{4}$$ because you have here:$\int_{-1}^{1} \frac{\arctan x}{\sqrt{1-x²}}=0$ as $\arctan$ is odd function over symetric domain.