Bài 3 (Trang 101 SGK Toán Giải tích 12):

Sử dụng phương pháp đổi biến số, hãy tính :

a) $\int {\left(1-x\right)}^9 dx$(đặt $u=1-x$) ;

b) $\int x{(1+x^2)}^{\frac{3}{2}}dx$ (đặt u = 1 + x²);

c) $\int {\cos }^{3}x\sin xdx$(đặt t = cos x);

d) $\int \frac{dx}{e^x +e^{-x}+2}$ (đặt u = e^x + 1).

Hướng dẫn giải:

a) Đặt u = 1 - x => du = -dx => dx = -du

$\int {\left(1-\mathrm{x}\right)}^9\mathrm{dx} =-\int {\mathrm{u}}^{9 }\mathrm{du} = -\frac{{\mathrm{u}}^{10}}{10}+\mathrm{C} =-\frac{1}{10}{\left(1-\mathrm{x}\right)}^{10}+\mathrm{C}$

b) Đặt $\mathrm{u}= 1+{\mathrm{x}}^2\Rightarrow \mathrm{du}=2\mathrm{xdx} \Rightarrow \mathrm{xdx}=\frac{1}{2}\mathrm{du}$

$\int \mathrm{x}{\left(1+{\mathrm{x}}^2\right)}^{\frac{3}{2}}\mathrm{dx}= \frac{1}{2}\int {\mathrm{u}}^{\frac{3}{2}}\mathrm{du}= \frac{1}{5}{\mathrm{u}}^{\frac{5}{2}}+ \mathrm{C} = \frac{1}{5}{\left(1+{\mathrm{x}}^2\right)}^{\frac{5}{2}} +\mathrm{C}$

c) Đặt t = cos x => dt = -sinxdx => sinxdx = -dt

$\int {\cos }^3\mathrm{xsinxdx} =-\int {\mathrm{t}}^3\mathrm{dt} = -\frac{{\mathrm{t}}^4}{4}+ \mathrm{C}=-\frac{1}{4}{\cos }^4\mathrm{x} + \mathrm{C}$

d) $\mathrm{I}=\int \frac{\mathrm{dx}}{{\mathrm{e}}^{\mathrm{x}}+ {\mathrm{e}}^{-\mathrm{x}}+2}= \int \frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}}{{\mathrm{e}}^{2\mathrm{x}}+2{\mathrm{e}}^{\mathrm{x}}+1 }=\int \frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}}{{\left({\mathrm{e}}^{\mathrm{x}}+1\right)}^2}$

Đặt t = e^x + 1 => dt = e^xdx

$\Rightarrow \mathrm{I}= \int \frac{\mathrm{dt}}{{\mathrm{t}}^2}=-\frac{1}{\mathrm{t}}+ \mathrm{C} =\frac{-1}{{\mathrm{e}}^{\mathrm{x}}+1 }+\mathrm{C}$