Verla
MathCalculus

Integral of Trig Inverse Functions: Evaluate an Arcsine Integral

Recognize the arcsine derivative in the denominator, substitute u = arcsin t, and obtain the antiderivative 1/2(arcsin t)^2 + C with its real-domain boundary.

Question

In the following exercise, compute the antiderivative using an appropriate substitution:

∫sin⁡−1t1−t2 dt.\int \frac{\sin^{-1} t}{\sqrt{1-t^2}}\,dt.

Answer

Here sin⁡−1t\sin^{-1}t means arcsin⁡t\arcsin t. Set

u=arcsin⁡t.u=\arcsin t.

Then

du=dt1−t2.du=\frac{dt}{\sqrt{1-t^2}}.

The integral becomes

∫u du=u22+C.\int u\,du=\frac{u^2}{2}+C.

Substituting back gives

12(arcsin⁡t)2+C.\boxed{\frac12\bigl(\arcsin t\bigr)^2+C}.

A derivative check confirms the result:

ddt[12(arcsin⁡t)2]=(arcsin⁡t)11−t2.\frac{d}{dt}\left[\frac12(\arcsin t)^2\right] =(\arcsin t)\frac{1}{\sqrt{1-t^2}}.

Evidence boundary

The notation follows OpenStax Exercise 411: sin⁻¹t is the principal inverse sine, not 1/sin t. For a real-valued antiderivative, the integrand is defined on −1<t<1; the endpoints make the derivative factor singular. The arbitrary constant C is required.

Sources

These references support the concepts and methods used in the explanation above.