Bài 1 (Trang 112 SGK Toán Giải tích 12):

Tính các tích phân sau:

a) ${\int }_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt[3]{{(1-\mathrm{x})}^2}\mathrm{dx}$;

b) ${\int }_0^{\frac{\mathrm{\pi }}{2}}\sin (\frac{\mathrm{\pi }}{4}-\mathrm{x})\mathrm{dx}$;

c) ${\int }_{\frac{1}{2}}^2\frac{1}{\mathrm{x}(\mathrm{x}+1)}\mathrm{dx};$

d) ${\int }_0^2\mathrm{x}{(\mathrm{x}+1)}^2\mathrm{dx}$;

e) ${\int }_{\frac{1}{2}}^2\frac{1-3\mathrm{x}}{{(\mathrm{x}+1)}^2}\mathrm{dx}$;

g)${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin 3\mathrm{xcos}5\mathrm{xdx}$.

Hướng dẫn giải:

a) Đặt $\mathrm{u}=1-\mathrm{x} \Rightarrow \mathrm{du}=-\mathrm{dx} \Rightarrow \mathrm{dx}=-\mathrm{du}$

Đổi cận:

Vậy: ${\int }_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt[3]{{(1-\mathrm{x})}^2}\mathrm{dx}=-{\int }_{\frac{1}{2}}^{\frac{3}{2}}{\mathrm{u}}^{\frac{2}{3}}\mathrm{du}=\frac{3}{5}{\mathrm{u}}^{\frac{5}{3}}{|}_{\frac{1}{2}}^{\frac{3}{2}} =\frac{3}{5}\mathrm{u}\sqrt[3]{{\mathrm{u}}^2}{|}_{\frac{1}{2}}^{\frac{3}{2}}=\frac{3}{5}(\frac{3}{2}\sqrt[3]{\frac{9}{4}} - \frac{1}{2}\sqrt[3]{\frac{1}{4}})$.

b) Ta có: ${\int }_0^{\frac{\mathrm{\pi }}{2}}\sin (\frac{\mathrm{\pi }}{4}-\mathrm{x})\mathrm{dx}=\cos (\frac{\mathrm{\pi }}{4}-\mathrm{x}){|}_0^{\frac{\mathrm{\pi }}{2}}=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2} = 0$

c) ${\int }_{\frac{1}{2}}^2\frac{1}{\mathrm{x}(\mathrm{x}+1)}\mathrm{dx}={\int }_{\frac{1}{2}}^2(\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1})\mathrm{dx}=\ln \left|\frac{\mathrm{x}}{\mathrm{x}+1}\right|{|}_{\frac{1}{2}}^2=\ln \frac{2}{3}-\ln \frac{1}{3}=\ln 2$

d) ${\int }_0^2\mathrm{x}{(\mathrm{x}+1)}^2\mathrm{dx}={\int }_0^2({\mathrm{x}}^3+2{\mathrm{x}}^2+\mathrm{x})\mathrm{dx}=(\frac{{\mathrm{x}}^4}{4}+\frac{2{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^2}{2}){|}_0^2=\frac{34}{3}$

e) Đặt $\mathrm{u}=\mathrm{x}+1 \Rightarrow \mathrm{du}=\mathrm{dx}.$

Đổi cận:

$$\begin{gathered} {\int }_{\frac{1}{2}}^2\frac{1-3x}{{(1+x)}^2}dx={\int }_{\frac{3}{2}}^3\frac{1-3(u-1)}{u^2}du={\int }_{\frac{3}{2}}^3(\frac{4}{u^2}-\frac{3}{u})du \\ (-\frac{4}{u}-3\ln \left|u\right|){|}_{\frac{3}{2}}^3=-\frac{4}{3}-3\ln 3+\frac{8}{3}+3\ln \frac{3}{2}=\frac{4}{3}-3\ln 2 \end{gathered}$$

g) ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin 3\mathrm{xcos}5\mathrm{xdx}=\frac{1}{2}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}(\sin 8\mathrm{x}-\sin 2\mathrm{x})\mathrm{dx}=(-\frac{1}{16}\cos 8\mathrm{x}+\frac{1}{4}\cos 2\mathrm{x}){|}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}=0.$