Nguyên hàm bài tập phần 7

\( \int \frac{\sin 2x \cdot \cos x}{1 + \cos x} \, dx \quad (2005B) \)

\( I = \int \frac{2 \sin x \cos^2 x}{1 + \cos x} \, dx \quad \)Đặt \( t = \cos x \)

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

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

Làm thêm

a) \( \int \frac{\cos x + \sin x \cos x}{2 + \sin x} \, dx \)

b) \( \int \frac{\sin 2x}{4 - \cos^2 x} \, dx \)

c) \( \int \frac{\tan x}{\cos^3 x} \, dx \)

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\( \int \frac{1}{\cos x} \, dx \)

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

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

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

Tương tự: \( \int \frac{1}{\sin x} \, dx \)

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\( \int \frac{\cos 3x}{\sin x} \, dx \)

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

\( = \int \sin x \frac{ (4 \cos^3 x - 3 \cos x)}{1 - \cos^2 x} \, dx \)

Đặt \( t = \cos x \)

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

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

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

\( = 2 \cos^2 x + \frac{1}{2} \ln |\cos^2 x - 1| + C \)

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\( \int \frac{\sin x + 2 \cos x}{\cos 2x} \, dx \)

Gợi ý:  

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

Đặt \( t = \cos x \) và \( t = \sin x \), tương ứng:  

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

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

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

\( = -\frac{\sqrt{2}}{4} \left[ \ln \left| \frac{t - \frac{\sqrt{2}}{2}}{ t + \frac{\sqrt{2}}{2}} \right| \right] + C \)

\( = -\frac{\sqrt{2}}{4} \ln \left| \frac{2 \cos x - \sqrt{2}}{2 \cos x + \sqrt{2}} \right| + C \)

* \( \int \frac{\cos x}{1 - 2 \sin^2 x} \, dx \): tương tự.  

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Gặp: \( \int \frac{1}{\cos^2 x} f(\tan x) \, dx \), đặt \( t = \tan x \)

 

\( \int \frac{1}{\cos^4 x} \, dx \)

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

\( = \tan x + \frac{\tan^3 x}{3} + C \)

Tổng quát!  
\( \int \frac{1}{\cos^{2n} x} \, dx \quad \text{và} \quad \int \frac{1}{\sin^{2n} x} \, dx \)

 

\( \int \frac{1 + \sin x}{\cos^4 x} \, dx \)

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

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