Integrals Involving Trig Functions — Question 2

Question 2 – Solution

Begin with the double-angle identity $$\sin(2x)=2\sin(x)\cos(x).$$ Squaring both sides gives $$\sin^2(x)\cos^2(x)=\frac14\sin^2(2x).$$ Now use $\sin^2(\theta)=\frac{1-\cos(2\theta)}2$ with $\theta=2x$: $$\sin^2(x)\cos^2(x)=\frac18\bigl(1-\cos(4x)\bigr).$$ Therefore, $$\begin{align*} \int_0^{\pi/2}\sin^2(x)\cos^2(x)\,dx &=\frac18\int_0^{\pi/2}\bigl(1-\cos(4x)\bigr)\,dx\\ &=\frac18\left[x-\frac14\sin(4x)\right]_0^{\pi/2}. \end{align*}$$ Evaluate the endpoints: $$\begin{align*} \frac18\left[\left(\frac\pi2-\frac14\sin(2\pi)\right) -\left(0-\frac14\sin(0)\right)\right] &=\frac18\left(\frac\pi2\right)\\ &=\frac\pi{16}. \end{align*}$$ Hence, $$\boxed{\displaystyle \int_0^{\pi/2}\sin^2(x)\cos^2(x)\,dx=\frac\pi{16}}.$$