Đáp án:
$\dfrac{1}{6}\tan x\sec^5x - \dfrac{7}{24}\tan x\sec^3x+\dfrac{1}{16}\tan x\sec x+ \dfrac{1}{16}\ln|\tan x + \sec x| + C$
Giải thích các bước giải:
$\begin{array}{l}\quad \displaystyle\int\dfrac{\tan^4x}{\cos^3x}dx\\ = \displaystyle\int \tan^4x.\sec^xdx\qquad với\,\,\sec x = \dfrac{1}{\cos x}\\ = \displaystyle\int (\sec^2x - 1)^2\sec^3xdx\\ = \displaystyle\int(\sec^7x - 2\sec^5x + \sec^3x)dx\\ = \displaystyle\int\sec^7xdx - 2\displaystyle\int\sec^5xdx + \displaystyle\int\sec^3xdx\\ = \left(\dfrac{\sin x\sec^6x}{6} + \dfrac{5}{6}\displaystyle\int\sec^5xdx \right) -2\left(\dfrac{\sin x\sec^4x}{4} + \dfrac{3}{4}\displaystyle\int\sec^3xdx \right) + \left(\dfrac{\sin x\sec^2x}{2} + \dfrac{1}{2}\displaystyle\int\sec xdx \right)\\ = \dfrac{1}{6}\tan x\sec^5x - \dfrac{7}{24}\tan x\sec^3x+\dfrac{1}{16}\tan x\sec x + \dfrac{1}{16}\displaystyle\int\sec xdx\\ = \dfrac{1}{6}\tan x\sec^5x - \dfrac{7}{24}\tan x\sec^3x+\dfrac{1}{16}\tan x\sec x+ \dfrac{1}{16}\ln|\tan x + \sec x| + C \end{array}$
