AP Calculus AB and BC

Integral of arcsin x / sqrt(1-x^2)

Q: How do you integrate arcsin x over sqrt(1-x^2)?

It is \frac{\left(\arcsin x\right)^{2}}{2} + C, from the substitution u = \arcsin x.

Q: Why is u = arcsin x rather than u = 1 - x^2?

Because du has to account for everything left in the integrand. With u = \arcsin x the entire factor \frac{dx}{\sqrt{1-x^{2}}} becomes du and nothing in x remains.

Q: Does this integral need integration by parts?

No. It is a one line substitution. Parts is for products of unrelated functions, not for a function sitting next to its own derivative.

The integral of arcsin x over the square root of one minus x squared is one half of the square of arcsin x, plus C. One over that square root is exactly the derivative of arcsin x, so u equals arcsin x absorbs the whole denominator into du and turns the problem into the integral of u du.

\int \frac{\arcsin x}{\sqrt{1-x^{2}}}\,dx = \frac{\left(\arcsin x\right)^{2}}{2} + C

Dividing by the root is multiplying by the derivative

The derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^{2}}}$ , and that factor is already sitting in the integrand. Set $u = \arcsin x$ , so $du = \frac{dx}{\sqrt{1-x^{2}}}$ and the whole denominator disappears into $du$ .

\int \frac{\arcsin x}{\sqrt{1-x^{2}}}\,dx = \int u\,du = \frac{u^{2}}{2} + C

= \frac{\left(\arcsin x\right)^{2}}{2} + C

Arcsine in the other role

Most students meet arcsine as an answer: it is the antiderivative of 1 over the square root of 1 minus x squared. Here it is the substitution instead. The same derivative fact does the work either way, and recognising it in both directions is the point of the problem.

Notation and domain

Write the result as $\frac{\left(\arcsin x\right)^{2}}{2}$ . The compressed form $\frac{\arcsin^{2}x}{2}$ is acceptable in most classes, but $\frac{\arcsin x^{2}}{2}$ is not: that reads as the arcsine of $x^{2}$ , a different function with a different derivative.

The integrand exists only on $-1 < x < 1$ , since $\sqrt{1-x^{2}}$ is zero at both endpoints. A definite integral running to $x = 1$ is improper and needs a limit, even though the antiderivative itself is perfectly well behaved there.

The mistakes students make

Two of these come from misreading the structure, and one from misplacing the square.

Substituting $u = 1-x^{2}$ out of habit. It matches the square root, but $du = -2x\,dx$ and there is no $x$ in the integrand to pay for it, so the substitution never closes.
Dropping the $\frac{1}{2}$ and answering $\left(\arcsin x\right)^{2} + C$ , whose derivative is twice the integrand.
Writing $\arcsin\left(x^{2}\right)$ or $\arcsin\left(\frac{x^{2}}{2}\right)$ , both of which square the input rather than the output.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

How do you integrate arcsin x over sqrt(1-x^2)?

It is $\frac{\left(\arcsin x\right)^{2}}{2} + C$ , from the substitution $u = \arcsin x$ .

Why is u = arcsin x rather than u = 1 - x^2?

Because $du$ has to account for everything left in the integrand. With $u = \arcsin x$ the entire factor $\frac{dx}{\sqrt{1-x^{2}}}$ becomes $du$ and nothing in $x$ remains.

Does this integral need integration by parts?

No. It is a one line substitution. Parts is for products of unrelated functions, not for a function sitting next to its own derivative.