Ví dụ 1: Tính nguyên hàm. b, ∫(2 sin x − 3 cos x) dx c, ∫(x+3)4dx d, ∫sin x . cos x3 dx​

Đáp án:

$\begin{array}{l}
b)\int {\left( {2\sin x - 3\cos x} \right)dx} \\
 = \int {2\sin xdx}  - \int {3\cos xdx} \\
 = 2.\cos x + 3\sin x + C\\
c)\int {\left( {x + 3} \right).4dx} \\
 = \int {\left( {4x + 12} \right)dx} \\
 = 2{x^2} + 12x + C\\
d)\int {\sin x.\cos x.3dx} \\
 = \int {\frac{3}{2}.2\sin x.\cos x.dx} \\
 = \frac{3}{2}.\int {\sin 2x.dx} \\
 = \frac{3}{4}.\int {\sin 2x.\left( {d2x} \right)} \\
 = \frac{3}{4}.\cos 2x + C
\end{array}$

$\begin{array}{l}a) \quad \displaystyle\int(2\sin x - 3\cos x)dx\\ = 2 \displaystyle\int\sin xdx - 3 \displaystyle\int\cos xdx\\ = 2.(-\cos x) - 3.\sin x + C\\ = -3\sin x - 2\cos x + C\qquad \text{(C: hằng số)}\\ b) \quad \displaystyle\int(x+3)^4dx\\ = \dfrac{(x+3)^{5}}{5} + C\qquad \text{(C: hằng số)}\\ c)\quad \displaystyle\int\sin x.\cos^3xdx\\ Đặt \,\,u = \cos x\\ \to du = -\sin xdx\\ \text{Ta được:}\\ -\displaystyle\int u^3du\\ = - \dfrac{u^4}{4} + C\\ = - \dfrac{1}{4}\cos^4x + C \qquad \text{(C: hằng số)}\\\end{array}$