Giải Bài 4.4 trang 12 SGK Toán 12

Lời giải

a) $\displaystyle\int \left(2\cos x - \dfrac{3}{\sin^2 x}\right)\mathrm{d}x$

Ta dùng $\displaystyle\int \cos x\,\mathrm{d}x = \sin x + C$ và $\displaystyle\int \dfrac{1}{\sin^2 x}\,\mathrm{d}x = -\cot x + C$.

$$\int \left(2\cos x - \dfrac{3}{\sin^2 x}\right)\mathrm{d}x = 2\int \cos x\,\mathrm{d}x - 3\int \dfrac{1}{\sin^2 x}\,\mathrm{d}x = 2\sin x + 3\cot x + C.$$

Kiểm tra: $\left(2\sin x + 3\cot x\right)' = 2\cos x - \dfrac{3}{\sin^2 x}$ (đúng).

b) $\displaystyle\int 4\sin^2 \dfrac{x}{2}\,\mathrm{d}x$

Hạ bậc: $\sin^2 \dfrac{x}{2} = \dfrac{1 - \cos x}{2}$, nên $4\sin^2 \dfrac{x}{2} = 2(1 - \cos x) = 2 - 2\cos x$.

$$\int 4\sin^2 \dfrac{x}{2}\,\mathrm{d}x = \int (2 - 2\cos x)\,\mathrm{d}x = 2x - 2\sin x + C.$$

Kiểm tra: $\left(2x - 2\sin x\right)' = 2 - 2\cos x = 4\sin^2 \dfrac{x}{2}$ (đúng).

c) $\displaystyle\int \left(\sin\dfrac{x}{2} - \cos\dfrac{x}{2}\right)^2 \mathrm{d}x$

Khai triển và dùng $\sin^2\dfrac{x}{2} + \cos^2\dfrac{x}{2} = 1$, $2\sin\dfrac{x}{2}\cos\dfrac{x}{2} = \sin x$:

$$\left(\sin\dfrac{x}{2} - \cos\dfrac{x}{2}\right)^2 = \sin^2\dfrac{x}{2} - 2\sin\dfrac{x}{2}\cos\dfrac{x}{2} + \cos^2\dfrac{x}{2} = 1 - \sin x.$$

Do đó:

$$\int \left(\sin\dfrac{x}{2} - \cos\dfrac{x}{2}\right)^2 \mathrm{d}x = \int (1 - \sin x)\,\mathrm{d}x = x + \cos x + C.$$

Kiểm tra: $\left(x + \cos x\right)' = 1 - \sin x$ (đúng).

d) $\displaystyle\int \left(x + \tan^2 x\right)\mathrm{d}x$

Dùng $\tan^2 x = \dfrac{1}{\cos^2 x} - 1$, nên $x + \tan^2 x = x + \dfrac{1}{\cos^2 x} - 1$.

$$\int \left(x + \tan^2 x\right)\mathrm{d}x = \int \left(x + \dfrac{1}{\cos^2 x} - 1\right)\mathrm{d}x = \dfrac{x^2}{2} + \tan x - x + C.$$

Kiểm tra: $\left(\dfrac{x^2}{2} + \tan x - x\right)' = x + \dfrac{1}{\cos^2 x} - 1 = x + \tan^2 x$ (đúng).

Đáp số