# Bài 50 Trang 176 SGK Đại số và Giải tích 12 Nâng cao

##### Hướng dẫn giải

a) Đặt

$$\left\{ \matrix{ u = {x^2} \hfill \cr dv = \sin 2xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = 2xdx \hfill \cr v = - {1 \over 2}\cos 2x \hfill \cr} \right.$$

Do đó $$\int\limits_0^{{\pi \over 2}} {{x^2}\sin 2xdx} = \left. { - {1 \over 2}{x^2}\cos 2x} \right|_0^{{\pi \over 2}} + \int\limits_0^{{\pi \over 2}} {{x^2}\cos 2xdx}$$
$$= {{{\pi ^2}} \over 8} + \int\limits_0^{{\pi \over 2}} {x\cos 2xdx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)}$$
Đặt

$$\left\{ \matrix{ u = x \hfill \cr dv = \cos 2xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = dx \hfill \cr v = {1 \over 2}\sin 2x \hfill \cr} \right.$$

Do đó $$\int\limits_0^{{\pi \over 2}} {x\cos 2xdx\, = \left. {{1 \over 2}x\sin 2x} \right|_0^{{\pi \over 2}}} - {1 \over 2}\int\limits_0^{{\pi \over 2}} {\sin 2xdx} = \left. {{1 \over 4}\cos 2x} \right|_0^{{\pi \over 2}} = - {1 \over 2}\,\,\,\,\,\,\,\,\left( 2 \right)$$
Thay (2) vào (1) ta được: $$\int\limits_0^{{\pi \over 2}} {{x^2}\sin 2xdx = {{{\pi ^2}} \over 8}} - {1 \over 2}.$$

b) Đặt $$u = 2{x^2} + 1 \Rightarrow du = 4xdx \Rightarrow xdx = {{du} \over 4}$$

$$\int\limits_1^2 {x\left( {2{x^2} + 1} \right)dx = {1 \over 4}} \int\limits_3^9 {udu} = \left. {{1 \over 8}{u^2}} \right|_3^9 = 9$$

c) Đặt $$u = {x^2} - 2x \Rightarrow du = 2\left( {x - 1} \right)dx \Rightarrow \left( {x - 1} \right)dx = {{du} \over 2}$$

$$\int\limits_2^3 {\left( {x - 1} \right)} {e^{{x^2} - 2x}}dx = {1 \over 2}\int\limits_0^3 {{e^u}du = } \left. {{1 \over 2}{e^u}} \right|_0^3 = {1 \over 2}\left( {{e^3} - 1} \right).$$