Evaluate the following integrals: ^1_0 ( ^ -1 x )^2dx

Question

Evaluate the following integrals:
$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}$

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Answer

We have,
$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}=(\cos^{-1}\text{x}\big)^2\int\limits^1_0\text{dx}-\int\limits^1_0\big(\int\text{dx}\big)\frac{\text{d}\big(\cos^{-1}\text{x}\big)^2}{\text{dx}}\text{ dx}$
$=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+\int^\limits{1}_0\frac{\text{x}\cdot2\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}$
Now,
Let $\cos^{-1}\text{x}=\text{t}\Rightarrow-\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$
When $\text{x}=0\Rightarrow\text{t}=\frac{\pi}{2}$
$\text{x}=1\Rightarrow\text{t}=0$
$\therefore\ \int^\limits{1}_0\frac{2\text{ x}\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}=-2\int^0_\limits{\frac{\pi}{2}}\text{t}\cos\text{t dt}=2\int^\limits{\frac{\pi}{2}}_0\text{t}\cos\text{t dt}$
$=2\Big[\text{t}\int\cos\text{t dt}-\int\big(\cos\text{t dt}\big)\frac{\text{dt}}{\text{dt}}\text{ dt}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\text{t}\sin\text{t}-\int\sin\text{t dt}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\text{t}\sin\text{t}+\cos\text{t}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\frac{\pi}{2}-1\Big]$
$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+\int^\limits{1}_0\frac{\text{x}\cdot2\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}\\=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+2\Big(\frac{\pi}{2}-1\Big)$
$=0-0+2\Big(\frac{\pi}{2}-1\Big)$
$=(\pi-2)$

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