$\displaystyle\int (\tan^7 x)(\sec^5 x)\ dx$.

$\tan^2 x = \sec^2 x - 1$

$I\ =\ \displaystyle\int (\tan^7 x)(\sec^5 x)\ dx = \displaystyle\int (\tan x)(\sec x)(\sec^2 x - 1)^3(\sec^4 x)\ dx$

Seja $u = \sec x$, $du = (\tan x)(\sec x) dx$.

$I\ =\ \displaystyle\int (u^2 - 1)^3 \cdot u^4\ du\ =\ \dfrac{u^{11}}{11} - \dfrac{u^9}{3} + \dfrac{3u^7}{7} - \dfrac{u^5}{5} + c$

$\fbox{$\displaystyle\int (\tan^7 x)(\sec^5 x)\ dx\ =\ \dfrac{\sec^{11} x}{11} - \dfrac{\sec^9 x}{3} + \dfrac{3\sec^7 x}{7} - \dfrac{\sec^5 x}{5} + c$}$