Integrals Involving Trig Functions — Question 3

Question 3 – Solution

We may use the interval $[0,2\pi]$, whose length is $2\pi$. The average-value formula is $$f_{\mathrm{avg}}=\frac1{2\pi}\int_0^{2\pi}\sin^4(x)\,dx.$$ Apply the power-reduction identity twice: $$\begin{align*} \sin^4(x) &=\left(\frac{1-\cos(2x)}2\right)^2\\ &=\frac14\left(1-2\cos(2x)+\cos^2(2x)\right)\\ &=\frac14\left(1-2\cos(2x)+\frac{1+\cos(4x)}2\right)\\ &=\frac{3-4\cos(2x)+\cos(4x)}8. \end{align*}$$ Thus, $$\begin{align*} f_{\mathrm{avg}} &=\frac1{2\pi}\int_0^{2\pi} \frac{3-4\cos(2x)+\cos(4x)}8\,dx\\ &=\frac1{16\pi} \left[3x-2\sin(2x)+\frac14\sin(4x)\right]_0^{2\pi}\\ &=\frac1{16\pi}(6\pi)=\frac38. \end{align*}$$ The sine terms vanish because their values are zero at both endpoints. Therefore, $$\boxed{f_{\mathrm{avg}}=\frac38}.$$