Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 14

Determine the integral $\displaystyle \int^{\pi}_0 \sin^2 t \cos^4 t dt$


$\begin{equation} \begin{aligned} \int^{\pi}_0 \sin^2 t \cos^4 t dt =& \int^{\pi}_0 \sin^2 t (\cos^2 t)^2 dt \qquad \text{Apply half-angle formulas } \cos 2 t = 1 - 2 \sin^2 t \text{ and } \cos 2t = 2 \cos^2 t - 1 \\ \\ \int^{\pi}_0 \sin^2 t \cos^4 t dt =& \int^{\pi}_0 \left( \frac{1 - \cos 2 t}{2} \right)\left( \frac{\cos 2 t + 1}{2} \right)^2 dt \\ \\ \int^{\pi}_0 \sin^2 t \cos^4 t dt =& \int^{\pi}_0 \left( \frac{1 - \cos 2t}{2}\right) \left( \frac{\cos^2 2 t + 2 \cos 2t + 1}{4} \right) dt \\ \\ \int^{\pi}_0 \sin^2 t \cos^4 t dt =& \int^{\pi}_0 \left( \frac{\cos^2 2t + 2 \cos 2 t + 1 - \cos^3 2t - 2 \cos^2 2t - \cos 2 t }{8} \right) dt \\ \ \int^{\pi}_0 \sin^2 t \cos^4 t dt =& \frac{1}{8} \int^{\pi}_0 (1 + \cos 2t - \cos^2 2t - \cos^3 2 t) dt \\ \\ \int^{\pi}_0 \sin^2 t \cos^4 t dt =& \frac{1}{8} \int^{\pi}_0 1 dt + \frac{1}{8} \int^{\pi}_0 \cos 2t dt - \frac{1}{8} \int^{\pi}_0 \cos^2 2t dt - \frac{1}{8} \int^{\pi}_0 \cos^3 2t dt \end{aligned} \end{equation}$


Let $u = 2t$, then $du = 2dt$, so $\displaystyle dt = \frac{du}{2}$. When $t = 0, u = 0$ and when $t = \pi, u = 2 \pi$. We integrate the equation term by term

@ 1st term


$\begin{equation} \begin{aligned} \frac{1}{8} \int^{\pi}_0 1 dt =& \frac{1}{8} \int^{2 \pi}_0 1 \cdot \frac{du}{2} \\ \\ \frac{1}{8} \int^{\pi}_0 1 dt =& \frac{1}{16} \int^{2 \pi}_0 1 du \\ \\ \frac{1}{8} \int^{\pi}_0 1 dt =& \frac{1}{16} \left[ u \right]^{2 \pi}_0 \\ \\ \frac{1}{8} \int^{\pi}_0 1 dt =& \frac{1}{16} (2 \pi - 0) \\ \\ \frac{1}{8} \int^{\pi}_0 1 dt =& \frac{2 \pi}{16} \\ \\ \frac{1}{8} \int^{\pi}_0 1 dt =& \frac{\pi}{8} \end{aligned} \end{equation}$


@ 2nd term


$\begin{equation} \begin{aligned} \frac{1}{8} \int^{\pi}_0 \cos 2t dt =& \frac{1}{8} \int^{2 \pi}_0 \cos u \cdot \frac{du}{2} \\ \\ \frac{1}{8} \int^{\pi}_0 \cos 2t dt =& \frac{1}{16} \int^{2 \pi}_0 \cos u du \\ \\ \frac{1}{8} \int^{\pi}_0 \cos 2t dt =& \frac{1}{16} \left[ \sin u \right]^{2 \pi}_0 \\ \\ \frac{1}{8} \int^{\pi}_0 \cos 2t dt =& \frac{1}{16} (\sin 2 \pi - \sin 0) \\ \\ \frac{1}{8} \int^{\pi}_0 \cos 2t dt =& \frac{1}{16} (0) \\ \\ \frac{1}{8} \int^{\pi}_0 \cos 2t dt =& 0 \end{aligned} \end{equation}$


@ 3rd term


$\begin{equation} \begin{aligned} \frac{1}{8} \int^{\pi}_0 \cos^2 2t dt =& \frac{1}{8} \int^{2 \pi}_0 \cos^2 u \cdot \frac{du}{2} \\ \\ \frac{1}{8} \int^{\pi}_0 \cos^2 2t dt =& \frac{1}{16} \int^{2 \pi}_0 \cos^2 u du \qquad \text{Apply half-angle formula } \cos 2 u = 2 \cos^2 u - 1 \\ \\ \frac{1}{8} \int^{\pi}_0 \cos^2 2t dt =& \frac{1}{16} \int^{2 \pi}_0 \left(\frac{\cos 2 u + 1}{2} \right) du \\ \\ \frac{1}{8} \int^{\pi}_0 \cos^2 2t dt =& \frac{1}{32} \int^{2 \pi}_0 (\cos 2u + 1) du \end{aligned} \end{equation}$


Let $v = 2u$, then $dv = 2 du$, so $\displaystyle du = \frac{dv}{2}$. When $u = 0, v = 0$ and when $u = 2 \pi, v = 4 \pi$


$\begin{equation} \begin{aligned} \frac{1}{32} \int^{32}_0 (\cos 2u + 1) du =& \frac{1}{32} \int^{4 \pi}_0 (\cos v + 1) \cdot \frac{dv}{2} \\ \\ \frac{1}{32} \int^{32}_0 (\cos 2u + 1) du =& \frac{1}{64} \int^{4 \pi}_0 (\cos v + 1) dv \\ \\ \frac{1}{32} \int^{32}_0 (\cos 2u + 1) du =& \frac{1 }{64} \left[ \sin v + v \right]^{4 \pi}_0 \\ \\ \frac{1}{32} \int^{32}_0 (\cos 2u + 1) du =& \frac{1}{64} (\sin 4 \pi + 4 \pi - \sin 0 - 0) \\ \\ \frac{1}{32} \int^{32}_0 (\cos 2u + 1) du =& \frac{1}{64} (0 + 4 \pi - 0 - 0) \\ \\ \frac{1}{32} \int^{32}_0 (\cos 2u + 1) du =& \frac{4 \pi}{64} \\ \\ \frac{1}{32} \int^{32}_0 (\cos 2u + 1) du =& \frac{\pi}{16} \end{aligned} \end{equation}$


@ 4th term


$\begin{equation} \begin{aligned} \frac{1}{8} \int^{\pi}_0 \cos^3 2t dt =& \frac{1}{8} \int^{2 \pi}_0 \cos^3 u \cdot \frac{du}{2} \\ \\ \frac{1}{8} \int^{\pi}_0 \cos^3 2t dt =& \frac{1}{16} \int^{2 \pi}_0 \cos^3 u du \\ \\ \frac{1}{8} \int^{\pi}_0 \cos^3 2t dt =& \frac{1}{16} \int^{2 \pi}_0 (\cos^2 u)(\cos u) du \qquad \text{Apply Trigonometric Identities } \cos^2 u + \sin^2 u = 1 \\ \\ \frac{1}{8} \int^{\pi}_0 \cos^3 2t dt =& \frac{1}{16} \int^{2 \pi}_0 (1 - \sin^2 u)(\cos u) du \end{aligned} \end{equation}$


Let $v = \sin u$, then $dv = \cos u du$. When $u = 0, v = 0$ and when $u = 2 \pi, v = 0$. Therefore,


$\begin{equation} \begin{aligned} \frac{1}{16} \int^{2 \pi}_0 (1 - \sin^2 u)(\cos u du) =& \frac{1}{16} \int^0_0 (1 - v^2) dv \\ \\ \frac{1}{16} \int^{2 \pi}_0 (1 - \sin^2 u)(\cos u du) =& \frac{1}{16} \left[ v - \frac{v^3}{3} \right]^0_0 \\ \\ \frac{1}{16} \int^{2 \pi}_0 (1 - \sin^2 u)(\cos u du) =& \frac{1}{16} (0) \\ \\ \frac{1}{16} \int^{2 \pi}_0 (1 - \sin^2 u)(\cos u du) =& 0 \end{aligned} \end{equation}$


Combine the results of integration term by term


$\begin{equation} \begin{aligned} \int^{\pi}_0 \sin^2 t \cos^4 t dt =& \frac{\pi}{8} + 0 - \frac{\pi}{16} - 0 \\ \\ \int^{\pi}_0 \sin^2 t \cos^4 t dt =& \frac{2 \pi + 0 - \pi - 0}{16} \\ \\ \int^{\pi}_0 \sin^2 t \cos^4 t dt =& \frac{\pi}{16} \end{aligned} \end{equation}$