Integral of Arcsine Function
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Theorem
- $\displaystyle \int \arcsin x \rd x = x \arcsin x + \sqrt {1 - x^2} + C$
for $x \in \left[{-1 \,.\,.\, 1}\right]$.
Proof
- $\displaystyle \int \arcsin x \rd x = \int 1 \cdot \arcsin x \rd x$
Using:
Expand on the above to demonstrate how
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we obtain:
- $\displaystyle \int \arcsin x \rd x = x \arcsin x - \int \dfrac x {\sqrt {1 - x^2}} \rd x$
| \(\displaystyle u\) | \(=\) | \(\displaystyle 1 - x^2\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \dfrac {\d u} {\d x}\) | \(=\) | \(\displaystyle -2 x\) | $\quad$ differentiate WRT $x$ | $\quad$ | |||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle -\frac 1 2 \frac {\d u} {\d x}\) | \(=\) | \(\displaystyle x\) | $\quad$ | $\quad$ | ||||||||
| \(\displaystyle \int \arcsin x \rd x\) | \(=\) | \(\displaystyle x \arcsin x - \int \frac {- \frac 1 2} {\sqrt u} \frac {\d u} {\d x} \rd x\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \int \arcsin x \rd x\) | \(=\) | \(\displaystyle x \arcsin x + \frac 1 2 \int u^{-1/2} \rd u\) | $\quad$ Integration by Substitution | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle x \arcsin x + u^{1/2} + C\) | $\quad$ Integral of Power | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle x \arcsin x + \sqrt {1-x^2} + C\) | $\quad$ from $u = 1 - x^2$ | $\quad$ |
$\blacksquare$
Sources
- Weisstein, Eric W. "Inverse Sine." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/InverseSine.html
