Nguyên hàm bài tập phần 3

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Nguyên hàm bài tập phần 3

Tính:

a) \(\int (2\cos 3x + 3\sin 2x) \, dx\)

b) \(\int \cos^2 x \, dx\)

\( = \int \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2} \left( x + \frac{1}{2} \sin 2x \right) + C\)

c) \(\int \sin^4 x \, dx\)

\(\sin^4 x = \left( \frac{1 - \cos 2x}{2} \right)^2 = \frac{1}{4} \left[1 + \cos^2 2x + 2\cos 2x \right]\)

\(= \frac{1}{4} \left[1 + \frac{1 + \cos 4x}{2} - 2\cos 2x \right]\)

\( = \frac{1}{8} \left( 3 + \cos 4x - 4\cos 2x \right)\)

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d) \( \int \cos^3 x \, dx \)

\( \cos 3x = 4\cos^3 x - 3\cos x \implies \cos^3 x = \frac{1}{4}\left[\cos 3x + 3\cos x\right] \)

\( \int \cos^3 x \, dx = \frac{1}{4} \int \left(\cos 3x + 3\cos x\right) dx \)

\( = \frac{1}{4} \left[\frac{1}{3} \sin 3x + 3\sin x \right] + C. \)

Cách 2: \( \int \cos^3 x \, dx = \int \cos x \left(1 - \sin^2 x\right) dx \)

\( = \int \cos x \, dx - \int \cos x \sin^2 x \, dx = \sin x - \frac{\sin^3 x}{3} + C. \)

e) \( \int \cos^5 x \, dx \)

\( \int \cos^5 x \, dx = \int \cos x \left(1 - \sin^2 x\right)^2 dx = \int \cos x \left(1 + \sin^4 x - 2\sin^2 x\right) dx \) \( = \sin x + \frac{\sin^5 x}{5} - 2\frac{\sin^3 x}{3} + C. \)

*Rút ra!

\(\int \sin^m x \, dx, \int \cos^m x \, dx\) (m chẵn: hạ bậc, m lẻ: đổi biến số)

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Tính: \( \int \cos 4x \sin 2x \, dx \quad \) (để lại trên bảng cho bài sau!)

\( \int \cos 4x \sin 2x \, dx = \frac{1}{2} \int \left[\sin 6x - \sin 2x\right] dx \)

\( = \frac{1}{2} \left[-\frac{1}{6} cos 6x + \frac{1}{2} \cos x\right] + C
= -\frac{1}{12} \cos 6x + \frac{1}{4} \cos 2x + C. \)

+ Rút ra: \( \int \cos ax \sin bx \, dx, \quad \int \cos ax \cos bx \, dx, \quad \int \sin ax \sin bx \, dx: \) Đưa về tổng

Làm thêm: \( \int \cos x \sin 2x \cos 4x \, dx \)

\( = \frac{1}{2} \int \cos \left(\sin 6x - \sin 2x\right) dx = \ldots \)

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Tính: \( \int \sin^4 x \cos^2 x \, dx \)

\( \cos^2 x \sin^4 x = \frac{(1 + \cos 2x)}{2} \frac{(3 + \cos 4x - 4\cos 2x)}{8} \)

\( = \frac{1}{16} \left[3 + \cos 4x - 4\cos 2x + 3\cos 2x + \cos 4x \cos 2x - 4\cos^2 2x \right] \)

\( = \frac{1}{16} \left[3 + \cos 4x - \cos 2x + \frac{1}{2} (\cos 6x + \cos 2x) - 4\frac{(1 + \cos 4x)}{2} \right] \)

\( = \frac{1}{32} \left[2 - 2\cos 4x - \cos 2x + \cos 6x\right] \)

\( \Rightarrow \int \cos^2 x \sin^4 x \, dx = \frac{1}{32} \left[2x - \frac{1}{2} \sin 4x - \frac{1}{2} \sin 2x + \frac{1}{6} \sin 6x\right] + C \)

b) \( \int \cos^3 x \sin^4 x \, dx \)

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

\( = \int \left(\cos x \sin^4 x - \cos x \sin^6 x\right) dx = \frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C. \)

*Nhớ!*

\( \int \sin^m x \cos^n x \, dx \)

- \(m, n\) đều chẵn: hạ bậc.

- \(m\) lẻ và \(n\) lẻ: đổi biến.

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

\( \int \frac{1}{\cos^2 u} \, du = \int (1 + \tan^2 u) \, du = \tan u + C \)

*\( \int \frac{1}{\sin^2 x} \, dx = \int (1 + \cot^2 x) \, dx = -\cot x + C \)

\( \int \frac{1}{\sin^2 u} \, du = \int (1 + \cot^2 u) \, du = -\cot u + C \)

Ví dụ:

1) \( \int \tan^2 x \, dx \)

2) \( \int (2\tan^2 x + 3\tan x - 4) \, dx \)

3) \( \int \tan^4 x \, dx \)

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

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

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