Integral of arcsin x: Answer, Parts Setup, Mistakes

How to integrate arcsin x by parts

The integrand $\arcsin x$ looks like a single function with nothing to substitute, which is exactly the signal for integration by parts. There is a second factor hiding in plain sight: the $dx$. Write the integrand as $\arcsin x \cdot 1$ and let that $1$ be the part you antidifferentiate.

Take $u = \arcsin x$, because differentiating it removes the inverse trig function entirely and leaves an algebraic expression. Take $dv = dx$, so $v = x$.

Substituting gives a new integral that is algebraic rather than inverse trigonometric, which is the whole point of the choice.

The leftover integral is a substitution. Let $w = 1 - x^{2}$, so $dw = -2x\,dx$ and $x\,dx = -\tfrac{1}{2}\,dw$.

Putting the two pieces together, the minus sign in front of the leftover integral meets the minus sign the substitution produced, and the result is a plus.

Why dv = dx is the right choice

Integration by parts is usually taught on products of two visible factors, so a lone $\arcsin x$ looks like the wrong tool. The reason it is the right tool is that parts does not need two factors, it needs a $u$ whose derivative is simpler and a $dv$ you can antidifferentiate. Here $dv = dx$ is trivially antidifferentiable and $\arcsin x$ becomes algebraic the moment you differentiate it.

The reverse choice does not work. There is no way to take $u = 1$, because then $du = 0$ and the formula collapses to nothing useful, and $dv = \arcsin x\,dx$ requires the very antiderivative you are trying to find.

The same trick handles the other single inverse functions on the course, which is worth recognizing as one pattern rather than three separate results.

• $\int \ln x\,dx = x\ln x - x + C$, with $u = \ln x$ and $dv = dx$
• $\int \arctan x\,dx = x\arctan x - \frac{1}{2}\ln\left(1 + x^{2}\right) + C$, same setup
• $\int \arcsin x\,dx = x\arcsin x + \sqrt{1 - x^{2}} + C$, same setup

In the LIATE ordering, Inverse trigonometric sits second, behind only Logarithmic. When an inverse trig function appears at all, it is almost always $u$.

Where the integral of arcsin x shows up on the AP exam

Integration by parts is Topic 6.11 in Unit 6 (Integration and Accumulation of Change), and the CED marks it BC only, so this integral can be asked on BC but not on AB. Unit 6 carries a weighting of 15 to 20 percent on both exams.

Be careful not to confuse this with the much more commonly tested integral that produces $\arcsin x$ rather than consuming it. That one is a basic antiderivative on both AB and BC.

A definite version is the usual exam form. The endpoints $0$ and $1$ are convenient because $\arcsin 1 = \frac{\pi}{2}$ and the square root vanishes there.

At $x = 1$ the bracket is $1 \cdot \frac{\pi}{2} + 0$, and at $x = 0$ it is $0 + 1$, so the difference is $\frac{\pi}{2} - 1$, roughly $0.571$.

Common mistakes with the integral of arcsin x

• Confusing the two directions. $\int \arcsin x\,dx$ is not $\frac{1}{\sqrt{1 - x^{2}}}$. That expression is the DERIVATIVE of $\arcsin x$, not its antiderivative.
• Writing $\frac{\left(\arcsin x\right)^{2}}{2}$. The power rule for antiderivatives applies to $x^{n}$, not to an arbitrary function, so this is only correct when the integrand also contains the derivative $\frac{1}{\sqrt{1 - x^{2}}}$.
• Getting the sign of the square root wrong. The substitution contributes a minus, and it meets the minus already in front of $\int v\,du$, so the term is $+\sqrt{1 - x^{2}}$, not $-\sqrt{1 - x^{2}}$.
• Forgetting the domain. $\arcsin x$ is only defined for $-1 \le x \le 1$, so a definite integral outside that interval is meaningless, not merely hard.
• Trying substitution first. Nothing inside $\arcsin x\,dx$ is the derivative of anything else in it, so there is no substitution to make.