Integral of arcsin(x)

6 Dec 2019 Leave a Comment

In this post I am going to show you how to integrate the inverse function of sin(x), arcsin(x), step-by-step. But first, let’s clear a few things.

What is arcsin(x)?

Arcsine is the inverse function of sine, therefore:

$\arcsin(\sin(x))=\sin(\arcsin(x))=x$

These are equivalent forms:

$\arcsin(x)=\text{a}\sin(x)=\sin^{-1}(x)$

The last form looks a lot like $x^{-1}$ , which is equal to 1/x. In the case of trigonometric functions though, it just means inverse function, not one over that function. So,

$\sin^{-1}(x)\neq \frac{1}{\sin(x)}$

but

$\sin^{-2}(x)= \left ( \sin(x) \right )^{-2}= \frac{1}{\left (\sin(x) \right )^2}=\frac{1}{\sin^2(x)}$

Anyway, we are now ready to integrate!

Integral of arcsin(x)

$\int \arcsin(x)\,\mathrm{d}x$

The technique required for this integral is integration by parts:

Since we want to solve the integral of arcsin(x), it would make no sense to let dv=arcsin(x)dx. Also, arcsin(x) is the same as $1\cdot \arcsin(x)$ , so we can integrate 1 and differentiate arcsin(x), whose derivative is $\frac{1}{\sqrt{1-x^2}}$ .

$u=\arcsin(x)\: \: \: \: \: dv=dx$

$du=\frac{1}{\sqrt{1-x^2}}dx\: \: \: \: \: v=x$

$x\arcsin(x)-\int \frac{x}{\sqrt{1-x^2}}dx$

Now we can use u-substitution, letting 1-x² be our u. When differentiating this, it will generate an x that cancels out with the one at the numerator; since we already used u in integration by parts, we will use t.

$t=1-x^2\: \: \: \: \: dt=-2x\,dx\rightarrow dx=-\frac{1}{2x}dt$