Calculus: The integral of the inverse trigonometric sine function.

Most books in Caculus do not treat the integrals of the
inverse trig and hyperbolic functions, but for completeness,
we include the derivation of the integrals of the inverse
trigonometric functions.

We derive the integral of the inverse sine (sin^{-1}, asin, arcsin) function.

Let y = sin^{-1} (x). Let us try integration by parts (IBP): \int udv = uv - \int v du.
Performing the substitution, u = sin^{-1}, then du = \frac{-2}{\sqrt{1-x^2}} dx.
Also let dv = dx, then v = x.

Applying IBP, we get,

\int sin^{-1} dx =   x sin^{-1} - \int x \frac{-2}{\sqrt{1-x^2}} dx

Consider the second term. Put u = \sqrt{1-x^2}. Then du = -2x dx.
The term then collapses to \int \frac{-2 x dx}{\sqrt{1-x^2}}, which is simply
the derivative of \sqrt{1-x^2}!

The integral of the inverse sine trigonometric sine function is:

\int sin^{-1}(x) = x \sin^{-1}(x)+ \sqrt{1-x^2} + C

.

The derivations of the integrals of the remaining inverse trig functions are done in the same manner.
Let's finish them up soon!

For more derivations of integrals and derivatives, consult Index to Calculus

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