Q. \(\int \sin \left(x^{2}\right)\,dx\)

We want to compute the integral

\[
\int \sin(x^2)\,dx.
\]

Step 1: Consider whether there is an elementary antiderivative.

When an integrand has the form \(\sin(x^2)\), a natural substitution would be \(u=x^2\), but that leads to a factor of \(dx\) that does not cancel neatly. The resulting integral is not elementary (it cannot be expressed using a finite combination of polynomials, exponentials, logarithms, and trigonometric functions).

Step 2: Use a standard special function approach.

There are special functions related to integrals of \(\sin(x^2)\) and \(\cos(x^2)\), known as the Fresnel sine integral.

The Fresnel sine integral is defined by

\[
S(z) = \int_{0}^{z} \sin\!\left(\frac{\pi}{2}t^2\right)\,dt.
\]

Step 3: Transform \(\sin(x^2)\) to match the Fresnel form.

We want to rewrite \(\sin(x^2)\) into a form like \(\sin\!\left(\frac{\pi}{2}t^2\right)\).

Let \(u = \sqrt{\tfrac{2}{\pi}}\,x\). Then

\[
x = \sqrt{\frac{\pi}{2}}\,u,
\quad
dx = \sqrt{\frac{\pi}{2}}\,du.
\]

Now compute the argument:

\[
x^2 = \left(\sqrt{\frac{\pi}{2}}\,u\right)^2 = \frac{\pi}{2}u^2.
\]

So the integrand becomes:

\[
\sin(x^2) = \sin\!\left(\frac{\pi}{2}u^2\right).
\]

Substitute into the integral:

\[
\int \sin(x^2)\,dx
=
\int \sin\!\left(\frac{\pi}{2}u^2\right)\left(\sqrt{\frac{\pi}{2}}\,du\right)
=
\sqrt{\frac{\pi}{2}}\int \sin\!\left(\frac{\pi}{2}u^2\right)\,du.
\]

Step 4: Express the remaining integral using the Fresnel sine integral.

By the definition of \(S(z)\), we have

\[
S(z) = \int_{0}^{z} \sin\!\left(\frac{\pi}{2}t^2\right)\,dt.
\]

That means an antiderivative of \(\sin\!\left(\frac{\pi}{2}u^2\right)\) is

\[
\int \sin\!\left(\frac{\pi}{2}u^2\right)\,du = S(u) + C.
\]

Therefore:

\[
\int \sin(x^2)\,dx
=
\sqrt{\frac{\pi}{2}}\,\Bigl(S(u)\Bigr) + C
=
\sqrt{\frac{\pi}{2}}\,S\!\left(\sqrt{\frac{2}{\pi}}\,x\right) + C.
\]

Final Answer

\[
\int \sin(x^2)\,dx
=
\sqrt{\frac{\pi}{2}}\,S\!\left(\sqrt{\frac{2}{\pi}}\,x\right) + C,
\]

where \(S(z)\) is the Fresnel sine integral.