Answer
$$1$$
Work Step by Step
\[
\int_{0}^{\sqrt{\pi}} x \sin x^{2} d x
\]
Substitute $u=x^{2}$, $du= 2 x d x \Longrightarrow x d x=\frac{1}{2} d u$
Substituting $0=x$ in $x^{2}=u$ gives us $0=u$
Substituting $\sqrt{\pi}=x$ in $x^{2}=u$ gives us $\pi=u$
Limits of integration will change from $\int_{0}^{\sqrt{\pi}}$ to $\int_{0}^{\pi}$
\[
=\frac{1}{2} \int_{0}^{\pi} \sin u d u=\frac{1}{2}[-\cos x]_{0}^{\pi}=2.\frac{1}{2} =1
\]
