How to calculate $E(\sin^2X)$

Question

If $X \sim N(0,1)$ then calculate $E(\sin^2X)$

I understand that $0 < \sin^2x<1$. So the expectation exists.

I proceed as

$E(\sin^2X)= \int_{-\infty}^{\infty}\sin^2xf(x)\,dx=2 \int_{0}^{\infty}\sin^2xf(x)\,dx$

Then I am unable to proceed.

Answer 1

It is a good idea to plug in $f(x)$.

$$\mathbb{E}[\sin^2 X]=\sqrt{\frac{2}{\pi}}\int_{0}^{+\infty}\sin^2(x) e^{-x^2/2}\,dx =\color{red}{\frac{\sinh 1}{\exp 1}}=0.432332358\ldots.$$ The middle equality follows from expanding $\sin^2(x)=\frac{1-\cos(2x)}{2}$ as a Taylor series, then integrating termwise. As an alternative, you may prove that: $$ \mathcal{L}\left(\frac{\cos x}{\sqrt{x}}\right) = \sqrt{\frac{\pi}{2}}\sqrt{\frac{s+\sqrt{1+s^2}}{1+s^2}}.$$