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