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You can work this out from basics as suggested by MathEagle, but if you are allowed to assume some basic knowledge of the Normal distribution there is a shortcut, which I think is probably what you are intended to do--
After you make the substitution, you should have
$\displaystyle \Gamma(1/2) = \sqrt{2} \; \int_0^\infty \exp(-x^2/2) \, dx$
which is equal to
$\displaystyle \sqrt{2} \cdot \sqrt{2 \pi} \cdot \frac{1}{\sqrt{2 \pi}} \; \int_0^{\infty} \exp(-x^2/2) \, dx$
Based on your knowledge of the Normal distribution, you should know the value of $\displaystyle \frac{1}{\sqrt{2 \pi}} \; \int_0^{\infty} \exp(-x^2/2) \, dx$.
Take it from there.