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

Determine the integral $\displaystyle \int^{\frac{\pi}{2}} \sin^2 (2 \theta) d \theta$


$
\begin{equation}
\begin{aligned}

\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \int^{\frac{\pi}{2}}_0 \left( \frac{1 - \cos (4 \theta)}{2} \right) d \theta
\qquad \text{Apply the half angle formula } \cos (2 \theta) = 1 - 2 \sin^2 (\theta)
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\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \frac{1}{2} \int^{\frac{\pi}{2}}_0 (1 - \cos 4 \theta) d \theta
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\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \frac{1}{2} \int^{\frac{\pi}{2}}_0 1 d \theta - \frac{1}{2} \int^{\frac{\pi}{2}}_0 \cos 4 \theta d \theta

\end{aligned}
\end{equation}
$


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


$
\begin{equation}
\begin{aligned}

\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \frac{1}{2} \int^{2 \pi}_0 1 \cdot \frac{du}{4} - \frac{1}{2} \int^{2 \pi}_0 \cos u \cdot \frac{du}{4}
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\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \frac{1}{8} \int^{2 \pi}_0 1 du - \frac{1}{8} \int^{2 \pi}_0 \cos u du
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\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \frac{1}{8} \cdot u \left. \right|^{2 \pi}_0 - \frac{1}{8} \cdot \sin u \left. \right|^{2 \pi}_0
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\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \frac{1}{8} \left[ u - \sin u \right]^{2 \pi}_0
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\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \frac{1}{8} [ 2 \pi - \sin (2 \pi) - 0 + \sin(0)]
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\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \frac{1}{8} (2 \pi - 0 - 0 + 0)
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\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \frac{1}{8} (2 \pi)
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\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \frac{2 \pi}{8}
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\int^{\frac{\pi}{2}}_0 \sin^2 (2 \theta) d \theta =& \frac{\pi}{4}

\end{aligned}
\end{equation}
$