Integral of arcsin(x)

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,

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

but

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

Anyway, we are now ready to integrate!

Integral of arcsin(x)

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

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 $1\cdot&space;\arcsin(x)$, so we can integrate 1 and differentiate arcsin(x), whose derivative is $\frac{1}{\sqrt{1-x^2}}$.

$u=\arcsin(x)\:&space;\:&space;\:&space;\:&space;\:&space;dv=dx$

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

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

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.

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

$x\arcsin(x)+\int&space;\frac{x}{\sqrt{t}}\frac{1}{2x}dt=x\arcsin(x)+\frac{1}{2}\int&space;\frac{1}{\sqrt{t}}dt=$

$x\arcsin(x)-\frac{1}{2}\int&space;t^{-\frac{1}{2}}dt=x\arcsin(x)-\frac{1}{2}2t^{\frac{1}{2}}=x\arcsin(x)-\sqrt{t}$

Let’s now undo the substitution (remember that t=1-x²):

$x\arcsin(x)+\sqrt{t}=x\arcsin(x)+\sqrt{1-x^2}$

Overall:

$\int&space;\arcsin(x)\,&space;\mathrm{d}x=x\arcsin(x)+\sqrt{1-x^2}+C$

And that’s it! Now try to solve this definite integral and leave the answer in the comments!

$\int_{-\frac{\sqrt3}{2}}^{\frac{\sqrt3}{2}}&space;\arcsin(x)\,&space;\mathrm{d}x$

I hope everything was clear and if you have any questions leave a comment and I’ll be happy to help!

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