Primitive of Square of Arcsine of x over a
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Theorem
- $\ds \int \paren {\arcsin \frac x a}^2 \rd x = x \paren {\arcsin \frac x a}^2 - 2 x + 2 \sqrt{a^2 - x^2} \arcsin \frac x a + C$
Proof
Let:
\(\ds u\) | \(=\) | \(\ds \arcsin \frac x a\) | ||||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \sin u\) | \(=\) | \(\ds \frac x a\) | Definition of Real Arcsine | |||||||||
\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \cos u\) | \(=\) | \(\ds \sqrt {1 - \frac {x^2} {a^2} }\) | Sum of Squares of Sine and Cosine |
Then:
\(\ds \int \paren {\arcsin \frac x a}^2 \rd x\) | \(=\) | \(\ds a \int u^2 \cos u \rd x\) | Primitive of Function of Arcsine | |||||||||||
\(\ds \) | \(=\) | \(\ds a \paren {2 u \cos u + \paren {u^2 - 2} \sin u} + C\) | Primitive of $x^2 \cos a x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a \paren {2 \arcsin \frac x a \sqrt {1 - \frac {x^2} {a^2} } + \paren {\paren {\arcsin \frac x a}^2 - 2} \frac x a} + C\) | substituting for $u$, $\sin u$ and $\cos u$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \paren {\arcsin \frac x a}^2 - 2 x + 2 \sqrt{a^2 - x^2} \arcsin \frac x a + C\) | simplifying |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.476$