Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 30

Determine the integral $\displaystyle \int^{\frac{\pi}{3}}_0 \tan^5 x \sec^6 x dx$


$
\begin{equation}
\begin{aligned}

\int^{\frac{\pi}{3}}_0 \tan^5 x \sec^6 x dx =& \int^{\frac{\pi}{3}}_0 \tan^5 x \sec^4 x \sec^2 x dx
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\int^{\frac{\pi}{3}}_0 \tan^5 x \sec^6 x dx =& \int^{\frac{\pi}{3}}_0 \tan^5 x (\sec^2 x)^2 \sec^2 x dx
\qquad \text{Apply Trigonometric Identity } \sec^2 x = \tan^2 x + 1
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\int^{\frac{\pi}{3}}_0 \tan^5 x \sec^6 x dx =& \int^{\frac{\pi}{3}}_0 \tan^ 2 x(\tan^2 x + 1)^2 \sec^2 x dx
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\int^{\frac{\pi}{3}}_0 \tan^5 x \sec^6 x dx =& \int^{\frac{\pi}{3}}_0 \tan^5x (\tan^4 + 2 \tan^2 x + 1) \sec^2x dx
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\int^{\frac{\pi}{3}}_0 \tan^5 x \sec^6 x dx =& \int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx

\end{aligned}
\end{equation}
$


Let $u = \tan x$, then $du = \sec^2 x dx$. When $x = 0, u = 0$ and when $\displaystyle x = \frac{\pi}{3}, u = \sqrt{3}$. Thus,


$
\begin{equation}
\begin{aligned}

\int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx =& \int^{\sqrt{3}}_0 (u^9 + 2u^7 + u^5) du
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\int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx =& \left[ \frac{u^{9 + 1}}{9 + 1} + 2 \left( \frac{u^{7 + 1}}{7 + 1} \right) + \frac{u^{5 + 1}}{5 + 1} \right]^{\sqrt{3}}_0
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\int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx =& \left[ \frac{u^{10}}{10} + \frac{2u^8}{8} + \frac{u^6}{6} \right]^{\sqrt{3}}_0
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\int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx =& \left[ \frac{u^{10}}{10} + \frac{u^8}{4} + \frac{u^6}{6} \right]^{\sqrt{3}}_0
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\int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx =& \frac{(\sqrt{3})^{10}}{10} + \frac{(\sqrt{3})^8}{4} + \frac{(\sqrt{3})^6}{6} - \frac{(0)^{10}}{10} - \frac{(0)^8}{4} - \frac{(0)^6}{6}
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\int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx =& \frac{[(\sqrt{3})^2]^5}{10} + \frac{[(\sqrt{3})^2]^4}{4} + \frac{[(\sqrt{3})^2]^3}{6}
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\int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx =& \frac{(3)^5}{10} + \frac{(3)^4}{4} + \frac{(3)^3}{6}
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\int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx =& \frac{243}{10} + \frac{81}{4} + \frac{27}{6}
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\int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx =& \frac{243}{10} + \frac{81}{4} + \frac{9}{2}
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\int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx =& \frac{486 + 405 + 90}{20}
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\int^{\frac{\pi}{3}}_0 (\tan^9 x + 2 \tan^7 x + \tan^5 x) \sec^2 x dx =& \frac{981}{20} = 49.05

\end{aligned}
\end{equation}
$