Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 4

Determine the integral $\displaystyle \int^{\frac{\pi}{2}}_0 \cos^5 x dx$


$
\begin{equation}
\begin{aligned}

\int^{\frac{\pi}{2}}_0 \cos^5 x dx =& \int^{\frac{\pi}{2}}_0 \cos^4 x \cos x dx
\qquad \text{Apply Trigonometric Identity } \cos^2x = 1 - \sin^2 x
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\int^{\frac{\pi}{2}}_0 \cos^5 x dx =& \int^{\frac{\pi}{2}}_0 (1 - sin^2 x)^2 \cos x dx
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\int^{\frac{\pi}{2}}_0 \cos^5 x dx =& \int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx

\end{aligned}
\end{equation}
$


Let $u = \sin x$, then $du = \cos x dx$, when $x = 0, u = 0$ and when $\displaystyle x = \frac{\pi}{2}, u = 1$. Therefore,


$
\begin{equation}
\begin{aligned}

\int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& \int^1_0 (1 - 2u^2 + u^4) du
\\
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\int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& u - 2 \cdot \frac{u^{2 + 1}}{2 + 1} + \left. \frac{u^{4 + 1}}{4 + 1} \right|^1_0
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\int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& u - \frac{2u^3}{3} + \frac{u^5}{5} \left. \right|^1_0
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\int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& 1 - \frac{2(1)^3}{3} + \frac{(1)^5}{5} - 0 + \frac{2(0)^3}{3} - \frac{(0)^5}{5}
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\int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& \frac{15 - 10 + 3}{15}
\\
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\int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& \frac{8}{15}


\end{aligned}
\end{equation}
$