How to integrate this integral?. How to integrate $$\int\limits_{1}^{3}\cos(5x^2)\,\mathrm dx \, ?$$ Doesn't seem to work by using intergation by parts or substitution.
 A: We can simply calculate the indefinite integral:
Substitute $u=\sqrt{\frac{10}{\pi}}x$ to get a Fresnel integral:
$$\int \cos (5x^2)\,\mathrm d x=\sqrt\frac{\pi}{10}\int\cos \left( \frac{\pi}{2}u^2\right)\,\mathrm d u=\sqrt\frac{\pi}{10} C(u)=\sqrt\frac{\pi}{10}\,C\left( \sqrt\frac{10}{\pi} x\right) +C$$
Now use the fundamental theorem of calculus to get:
$$\int_1^3 \cos (5x^2)\,\mathrm d x=\sqrt\frac{\pi}{10}\,C\left( 3\sqrt\frac{10}{\pi}\right)-\sqrt\frac{\pi}{10}\,C\left( \sqrt\frac{10}{\pi} \right)=\\
\boxed{\sqrt\frac{\pi}{10}\left( C\left(3\sqrt\frac{10}{\pi}\right)-C\left(\sqrt\frac{10}{\pi}\right)\right)}\approx 0.124$$

In general, we have:
$$ \int \cos (ax^2)\,\mathrm d x=\sqrt\frac{\pi}{2a}C\left( \sqrt\frac{2a}{\pi}x\right) + C$$
and thus
$$\int_b^c \cos (ax^2)\,\mathrm d x=\sqrt\frac{\pi}{2a}\left( C\left(c\sqrt\frac{2a}{\pi}\right)-C\left(b\sqrt\frac{2a}{\pi}\right)\right)$$
This identity can be derived analogically by substituting $u=\sqrt\frac{2a}{\pi}x$
A: $$\mathcal{I}=\int_1^3\cos\left(5x^2\right)\space\text{d}x=$$

Sustitute $u=\frac{\sqrt{2}\sqrt{5}x}{\sqrt{\pi}}$ and $\text{d}u=\frac{\sqrt{2}\sqrt{5}}{\sqrt{\pi}}\space\text{d}x$.
This gives a new lower bound $u=\frac{\sqrt{2}\sqrt{5}}{\sqrt{\pi}}$ and upper bound $u=\frac{3\sqrt{2}\sqrt{5}}{\sqrt{\pi}}$:

$$\mathcal{I}=\frac{\sqrt{\pi}}{\sqrt{2}\sqrt{5}}\int_{\frac{\sqrt{2}\sqrt{5}}{\sqrt{\pi}}}^{\frac{3\sqrt{2}\sqrt{5}}{\sqrt{\pi}}}\cos\left(\frac{\pi u^2}{2}\right)\space\text{d}u$$
his is a special integral, called the Fresnel integral.

Another way, is using series:
$$\cos\left(5x^2\right)=\sum_{n=0}^\infty\frac{(-25)^n(x^2)^{2n}}{(2n)!}$$
So, we get:
$$\mathcal{I}=\int_1^3\cos\left(5x^2\right)\space\text{d}x=\sum_{n=0}^\infty\frac{(-25)^n}{(2n)!}\int_1^3(x^2)^{2n}\space\text{d}x=\sum_{n=0}^\infty\frac{(-25)^n}{(2n)!}\cdot\frac{3^{1+4n}-1}{1+4n}$$
