Evaluate the following integrals: _ 0 ^ 1 2 x ^ -1 x 1-x^2 dx

Question

Evaluate the following integrals:
$\int_{0}^\limits{\frac{1}{2}}\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{ dx}$

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Answer

Let $\text{I}=\int_{0}^\limits{\frac{1}{2}}\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{ dx}$
Put $\text{x}=\sin\theta$
$\therefore\text{ dx}=\cos\theta\text{ d}\theta$
When $\text{x}\rightarrow0,\theta\rightarrow0$
When $\text{x}\rightarrow\frac{1}{2},\theta\rightarrow\frac{\pi}{6}$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{6}}\frac{\sin\theta\sin^{-1}(\sin\theta)}{\cos\theta}\cos\theta\text{ d}\theta$
$=\int_{0}^\limits{\frac{\pi}{6}}\theta\sin\theta\text{ d}\theta$
Applying integration by parts, we have
$\text{I}=\big[\theta\big(-\cos\theta\big)\big]^{\frac{\pi}{6}}_0-\int_{0}^\limits{\frac{\pi}{6}}1\times\big(-\cos\theta\big)\text{d}\theta$
$=-\Big(\frac{\pi}{6}\cos\frac{\pi}{6}-0\Big)+\int_{0}^\limits{\frac{\pi}{6}}\cos\theta\text{ d}\theta$
$=-\frac{\pi}{6}\times\frac{\sqrt{3}}{2}+\big[\sin\theta\big]^{\frac{\pi}{6}}_0$
$=-\frac{\pi}{4\sqrt{3}}+\Big(\sin\frac{\pi}{6}-\sin0\Big)$
$=-\frac{\pi}{4\sqrt{3}}+\Big(\frac{1}{2}-0\Big)$
$=\frac{1}{2}-\frac{\pi}{4\sqrt{3}}$

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