# Integral of S(x) and C(x)

Here I talked about special functions, and at the end of the post I said I would explain how to integrate these special functions. So here I am, showing you how to integrate the functions S(x) and C(x). If you gave it a try, let me know in the comments if you used this approach!

Integral of S(x)

$\int&space;\text{S}(x)\,&space;dx$

Like we did for the previous integrals, we are going to use integration by parts, as it is the only way to go.

$u=&space;\text{S}(x)\;\;\;\;\;dv=dx$ $du=&space;\sin(x^2)\,&space;dx\;\;\;\;\;v=x$ $x\text{S}(x)-\int&space;x\sin(x^2)dx$

Now we can use u-substitution to solve the integral:

$u=x^2\;&space;\;&space;\;&space;\;&space;\;&space;du=2x\,&space;dx\rightarrow&space;dx=\frac{1}{2x}du$ $\int&space;x\sin(u)\frac{1}{2x}du=\frac{1}{2}\int&space;\sin(u)du=-\frac{1}{2}\cos(u)\overset{x}{\rightarrow}-\frac{1}{2}\cos(x^2)$

Overall:

$\int&space;\text{S}(x)\,&space;dx=x\text{S}(x)+\frac{1}{2}\cos(x^2)+C$

Integral of C(x)

Believe me when I say that the procedure for this integral is exactly the same as that of the integral of S(x)! The only difference is that now we have the function cosine, and as you know the integral of cos(x) is sin(x). This is why the result is

$\int&space;\text{C}(x)\,&space;dx=x\text{C}(x)-\frac{1}{2}\sin(x^2)+C$

You can try this yourself and let me know if you managed to get the same result!

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