\( \int \frac{\sin^2 x}{\cos^4 x} \, dx = \int \frac{1}{\cos^2 x} \cdot \tan^2 x \, dx = \frac{\tan^3 x}{3} + C \)

Cách 2

\( I = \int \frac{1 - \cos^2 x}{\cos^4 x} \, dx = \int \frac{1}{\cos^4 x} \, dx - \int \frac{1}{\cos^2 x} \, dx \)

page40

\(I = \int \frac{1}{\cos^2 x}\cdot \frac{1}{ (\tan^2 x - 3\tan x - 4)} \, dx \quad \text{đặt } t = \tan x\)

\(= \int \frac{1}{t^2 - 3t - 4} \, dt = \int \frac{1}{(t + 1)(t - 4)} \, dt\)

\(= \frac{1}{5} \int \left( \frac{1}{t - 4} - \frac{1}{t + 1} \right) \, dt = \frac{1}{5} \ln \left| \frac{t - 4}{t + 1} \right| + C\)

page41 

\(\int \frac{1}{\cos 2x} \, dx = \int \frac{1}{\cos^2 x - \sin^2 x} \, dx\)

\(=\int \frac{1}{\cos^2 x (1 - \tan^2 x)} \, dx \quad \text{đặt } t = \tan x\)

\(=\int \frac{1}{1 - t^2} \, dt = \int \frac{1}{(1 - t)(1 + t)} \, dt\)

\(=\frac{1}{2} \int \left( \frac{1}{1 - t} + \frac{1}{1 + t} \right) \, dt\)

\(=\frac{1}{2} \ln \left| \frac{t + 1}{1 - t} \right| + C\)

page42

\(\int \frac{\tan^4 x}{\cos^2 x - \sin^2 x} \, dx = \int \frac{\tan^4 x}{\cos^2 x (1 - \tan^2 x)} \, dx \quad \text{đặt } t = \tan x\)

\(I = \int \frac{t^4}{1 - t^2} \, dt = \int \frac{t^4 - 1 + 1}{1 - t^2} \, dt = \int \left( -t^2 - 1 + \frac{1}{1 - t^2} \right) \, dt\)

\(= -\frac{t^3}{3} - t + \int \frac{1}{(1 - t)(1 + t)} \, dt\)

\(= -\frac{t^3}{3} - t + \frac{1}{2} \int \left( \frac{1}{1 - t} + \frac{1}{1 + t} \right) \, dt\)

\(= -\frac{t^3}{3} - t + \frac{1}{2} \ln \left| \frac{1 + t}{1 - t} \right| + C\)

page43

\(\int \frac{1}{\sin x \cos^3 x} \, dx = \int \frac{1}{\cos^4 x \cdot \tan x} \, dx\)

\(= \int \frac{1}{\cos^2 x} \cdot \frac{1 + \tan^2 x}{\tan x} \, dx\)

\(= \int \frac{1}{\cos^2 x} \cdot \frac{1}{\tan x} \, dx + \int \frac{1}{\cos^2 x} \cdot \tan x \, dx\)

\(= \ln |\tan x| + \frac{\tan^2 x}{2} + c.\)

page44