Primitive of Arcsine Function

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Theorem

$\ds \int \arcsin x \rd x = x \arcsin x + \sqrt {1 - x^2} + C$


Corollary

$\ds \int \arcsin \frac x a \rd x = x \arcsin \frac x a + \sqrt {a^2 - x^2} + C$


Proof 1

Let:

\(\ds u\) \(=\) \(\ds \arcsin x\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \sin u\) \(=\) \(\ds x\) Definition of Real Arcsine
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \cos u\) \(=\) \(\ds \sqrt {1 - x^2}\) Sum of Squares of Sine and Cosine


Then:

\(\ds \int \arcsin x \rd x\) \(=\) \(\ds \int u \cos u \rd u\) Primitive of Function of Arcsine
\(\ds \) \(=\) \(\ds \cos u + u \sin u + C\) Primitive of $x \cos a x$
\(\ds \) \(=\) \(\ds \cos u + u x + C\) Substitution for $\sin u$ from $\paren 1$
\(\ds \) \(=\) \(\ds \sqrt {1 - x^2} + u x + C\) Substitution for $\cos u$ from $\paren 2$
\(\ds \) \(=\) \(\ds x \arcsin x + \sqrt {1 - x^2} + C\) Substitution for $u$ and rearranging

$\blacksquare$


Proof 2

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \arcsin x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac 1 {\sqrt {1 - x^2} }\) Derivative of $\arcsin x$


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds x\) Primitive of Constant


Then:

\(\ds \int \arcsin x \rd x\) \(=\) \(\ds x \arcsin x - \int x \paren {\frac 1 {\sqrt {1 - x^2} } } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds x \arcsin x - \int \frac {x \rd x} {\sqrt {1 - x^2} } + C\) simplifying
\(\ds \) \(=\) \(\ds x \arcsin x - \paren {-\sqrt {1 - x^2} } + C\) Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$, with $a := 1$
\(\ds \) \(=\) \(\ds x \arcsin x + \sqrt {1 - x^2} + C\) simplifying

$\blacksquare$


Also see


Sources

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