Q. \[ \int \cos\left(x^2\right)\,dx \]

We want to evaluate the indefinite integral

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

Step 1: Recognize there is no simple elementary antiderivative.

The function \(\cos(x^2)\) has a quadratic inside the cosine. Integrals of the form \(\cos(ax^2)\) and \(\sin(ax^2)\) generally do not have antiderivatives expressible using elementary functions (polynomials, exponentials, logs, trig, and their combinations).

Step 2: Use the Fresnel cosine integral.

The standard special function used for this case is the Fresnel cosine integral, defined by

\[
C(u)=\int_{0}^{u} \cos\!\left(\frac{\pi}{2}\,t^2\right)\,dt.
\]

Step 3: Convert \(\cos(x^2)\) into \(\cos\!\left(\frac{\pi}{2}t^2\right)\) by substitution.

We rewrite \(x^2\) as a constant times \(t^2\). Let

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

Then

\[
x=\sqrt{\frac{\pi}{2}}\,u
\]

and

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

Now substitute into the integrand:

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

So the integral becomes

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

Step 4: Match the integral with the definition of \(C(u)\).

From the definition, we know that

\[
C(u)=\int_{0}^{u} \cos\!\left(\frac{\pi}{2}\,t^2\right)\,dt.
\]

Therefore an antiderivative of \(\cos\!\left(\frac{\pi}{2}\,u^2\right)\) is \(C(u)\), up to a constant. Concretely,

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

Thus

\[
\int \cos(x^2)\,dx
=\sqrt{\frac{\pi}{2}}\,C(u)+\text{constant}.
\]

Step 5: Substitute back \(u=\sqrt{\frac{2}{\pi}}\,x\).

\[
\int \cos(x^2)\,dx
=\sqrt{\frac{\pi}{2}}\,C\!\left(\sqrt{\frac{2}{\pi}}\,x\right)+C_0,
\]
where \(C_0\) is an arbitrary constant.

Final Answer:

\[
\boxed{\int \cos(x^2)\,dx=\sqrt{\frac{\pi}{2}}\,C\!\left(\sqrt{\frac{2}{\pi}}\,x\right)+C_0.}
\]