Answer
$\frac{sin^3x}{3} - \frac{sin^5x}{5} +C$
Work Step by Step
Find the indefinite integral
$\int \frac{tan^2x}{sec^5x}dx$, Convert tangents to secants
$\int \frac{sec^2x -1}{sec^5x}dx$
$\int[\frac{1}{sec^3x}- \frac{1}{sec^5x}]dx$
$\int[cos^3x- cos^5x]dx$
$\int cos^3xdx- \int cos^5xdx$, Break into two integrals
$\int (1-sin^2x)cosxdx- \int (1-sin^2x)^2 cosxdx$
$\int (1-sin^2x)cosxdx - \int (1-2sin^2x+sin^4x)cosxdx$
Use u-substitution, let $u=sinx$, $du=cosxdx$
$\int (1-u^2)du - \int (1-2u^2 +u^4)du$, Integrate
$-\frac{u^3}{3}+ \frac{2u^3}{3}- \frac{u^5}{5} +C$
$\frac{sin^3x}{3} - \frac{sin^5x}{5} +C$, Resubstitute
