Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 22

Determine the integral $\displaystyle \int^{\frac{\pi}{2}}_0 \sec^4 \left( \frac{t}{2} \right) dt$

Let $\displaystyle u = \frac{t}{2}$, then $\displaystyle du = \frac{1}{2} dt$, so $dt = 2 du$. When $t = 0,u = 0$ and when $\displaystyle t = \frac{\pi}{2}, u = \frac{\pi}{4}$


$
\begin{equation}
\begin{aligned}

\int^{\frac{\pi}{2}}_0 \sec^4 \left( \frac{t}{2} \right) dt =& \int^{\frac{\pi}{4}}_0 \sec^4 u \cdot 2 du
\\
\\
\int^{\frac{\pi}{2}}_0 \sec^4 \left( \frac{t}{2} \right) dt =& 2 \int^{\frac{\pi}{4}}_0 \sec^4 u du
\\
\\
\int^{\frac{\pi}{2}}_0 \sec^4 \left( \frac{t}{2} \right) dt =& 2 \int^{\frac{\pi}{4}}_0 \sec^2 u \sec^2 u du \qquad \text{Apply Pythagorean Identity } \sec^2 u = \tan^2 u + 1
\\
\\
\int^{\frac{\pi}{2}}_0 \sec^4 \left( \frac{t}{2} \right) dt =& 2 \int^{\frac{\pi}{4}}_0 (\tan^2 u + 1) \sec^2 u du

\end{aligned}
\end{equation}
$


Let $v = \tan u$, then $dv = \sec^2 u du$. When $u = 0, v = 0$ and when $\displaystyle u = \frac{\pi}{4}, v = 1$. Thus,


$
\begin{equation}
\begin{aligned}

2 \int^{\frac{\pi}{4}}_0 (\tan^2 u + 1) \sec^2 u du =& 2 \int^2_1 (v^2 + 1) dv
\\
\\
2 \int^{\frac{\pi}{4}}_0 (\tan^2 u + 1) \sec^2 u du =& 2 \left[ \frac{v^{2 + 1}}{2+ 1} + v \right]^1_0
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2 \int^{\frac{\pi}{4}}_0 (\tan^2 u + 1) \sec^2 u du =& 2 \left[ \frac{v^3}{3} + v \right]^1_0
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2 \int^{\frac{\pi}{4}}_0 (\tan^2 u + 1) \sec^2 u du =& 2 \left[ \frac{(1)^3}{3} + 1 - \frac{(0)^3}{3} - 0 \right]
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2 \int^{\frac{\pi}{4}}_0 (\tan^2 u + 1) \sec^2 u du =& 2 \left( \frac{1}{3} + 1 \right)
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2 \int^{\frac{\pi}{4}}_0 (\tan^2 u + 1) \sec^2 u du =& 2 \left( \frac{4}{3} \right)
\\
\\
2 \int^{\frac{\pi}{4}}_0 (\tan^2 u + 1) \sec^2 u du =& \frac{8}{3}

\end{aligned}
\end{equation}
$