Bài 3 (Trang 113 SGK Toán Giải tích 12):

Sử dụng phương pháp biến đổi số, tính tích phân sau

a) ${\int }_0^3\frac{{\mathrm{x}}^2}{{(1-\mathrm{x})}^{{\displaystyle \frac{3}{2}}}}\mathrm{dx}$ (đặt u = x + 1);

b) ${\int }_0^1\sqrt{1-{\mathrm{x}}^2}\mathrm{dx}$  (đặt x = sin t);

c) ${\int }_0^1\frac{{\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})}{1+{\mathrm{xe}}^{\mathrm{x}}}\mathrm{dx}$  (đặtu = 1 + x.e^x);

d) ${\int }_0^{\frac{\mathrm{a}}{2}}\frac{1}{\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}\mathrm{dx}$ (a>0) (đặt x=asin t).

Hướng dẫn giải:

a) Đặt u = x + 1 => du = dx

$$\begin{gathered} {\int }_0^3\frac{{\mathrm{x}}^2\mathrm{dx}}{(1+\mathrm{x}){\displaystyle \frac{3}{2}}} = {\int }_1^4\frac{{(\mathrm{u}-1)}^2}{{\mathrm{u}}^{{\displaystyle \frac{3}{2}}}}\mathrm{du} = {\int }_1^4\frac{{\mathrm{u}}^2-2\mathrm{u}+1}{{\mathrm{u}}^{{\displaystyle \frac{3}{2}}}}\mathrm{du} = {\int }_1^4({\mathrm{u}}^2-2{\mathrm{u}}^{-\frac{1}{2}}+{\mathrm{u}}^{-\frac{3}{2}})\mathrm{du} \\ = (\frac{2}{3}{\mathrm{u}}^{\frac{3}{2}}-4{\mathrm{u}}^{\frac{1}{2}}-2{\mathrm{u}}^{-\frac{1}{2}}){|}_1^4 \\ = \frac{5}{3} \end{gathered}$$

b) Đặt x = sin t => dx = costdt

Đổi cận:

$$\begin{gathered} {\int }_0^1\sqrt{1-{\mathrm{x}}^2}\mathrm{dx} = {\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{1-{\sin }^2\mathrm{t}}.\mathrm{costdt} = {\int }_0^{\frac{\mathrm{\pi }}{2}}{\cos }^2\mathrm{tdt} = \frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}(1+\cos 2\mathrm{t})\mathrm{dt} \\ = \frac{1}{2}(\mathrm{t}+\frac{1}{2}\sin 2\mathrm{t}){|}_0^{\frac{\mathrm{\pi }}{2}} = \frac{\mathrm{\pi }}{4}. \end{gathered}$$

c) Đặt u = 1 + xe^x => du = (e^x + xe^x)dx = e^x(1+x)dx

Đổi cận:

${\int }_0^1\frac{{\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})}{1+{\mathrm{xe}}^{\mathrm{x}}}\mathrm{dx} = {\int }_1^{1+\mathrm{e}}\frac{\mathrm{du}}{\mathrm{u}}=\ln \left|\mathrm{u}\right|{|}_1^{1+\mathrm{e}} = \ln (1+\mathrm{e})$

d) Đặt x = a.sint t => dx = a.costdt

Đổi cận:

${\int }_0^{\frac{\mathrm{a}}{2}}\frac{1}{\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}\mathrm{dx} = {\int }_0^{\frac{\mathrm{\pi }}{6}}\frac{\mathrm{acostdt}}{\sqrt{{\mathrm{a}}^2(1-{\sin }^2\mathrm{t})}}\mathrm{dt} = {\int }_0^{\frac{\mathrm{\pi }}{6}}\mathrm{dt} = \mathrm{t}{|}_0^{\frac{\mathrm{\pi }}{6}} = \frac{\mathrm{\pi }}{6}$.