# Integral of arccos(x)

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

What is arccos(x)?

Arccosine is the inverse function of cosine, therefore:

$\arccos(\cos(x))=\cos(\arccos(x))=x$

These are equivalent forms:

$\arccos(x)=\text{a}\cos(x)=\cos^{-1}(x)$

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,

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

but

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

Anyway, we are now ready to integrate!

Integral of arccos(x)

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

The technique required for this integral is integration by parts:

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

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

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

$x\arccos(x)+\int\frac{x}{\sqrt{1-x^2}}\,dx$

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.

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

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

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

Overall:

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

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

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

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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