Answer to Question #108091 in Calculus for Nimra

Question

∫ (x)= ∫ sinx²y dx

Answer 1

Let $I=\int sinx^2y dx$

Here y is constant ,Since integration is with respect to x .

$\implies I=\int sin (x\sqrt{y})^2.$

Let $x\sqrt{y}=u$ $\implies dx=\frac{1}{\sqrt{y}} du.$

$\implies I=\frac{1}{\sqrt{y}}\int sinu^2 du.$

By fundamental theorem of calculus ,every continuous real function f has an anti derivative but some ordinary looking continuous function have anti derivative that can not expressed in simple formula using more basic function. $I$ looks like one of them.

$I$ is called Fresnel integral which is a transcendental function.

$I$ admits the following power series expansion that converges for all $x$ .

$$I=\frac{1}{\sqrt{y}}\int \sum_{n=0}^{n=\infin} \frac{ (-1)^n}{(2n+1)!}u^{2(2n+1)} du$$

Where

$$\text{sin}u=\sum_{n=0}^{n=\infin} \frac{(-1)^n}{(2n+1)!} u^{2n+1}$$

$$\implies I=\frac{1}{\sqrt{y}} \sum_{n=0}^{n=\infin} \frac{(-1)^n}{(2n+1)! (4n+3)}u^{4n+3}$$

$$\implies I=\frac{1}{\sqrt{y}} \sum_{n=0}^{n=\infin}\frac{(-1)^n}{(2n+1)!(4n+3)}x^{4n+3}y^{\frac{1}{2} (4n+3)}$$