# Bài 18 Trang 161 SGK Đại số và Giải tích 12 Nâng cao

##### Hướng dẫn giải

a) Đặt

$$\left\{ \matrix{ u = \ln x \hfill \cr dv = {x^5}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = {{dx} \over x} \hfill \cr v = {{{x^6}} \over 6} \hfill \cr} \right.$$

$$\int\limits_1^2 {{x^5}} \ln xdx = \left. {{{{x^6}} \over 6}\ln x} \right|_1^2 - {1 \over 6}\int\limits_1^2 {{x^5}} dx = \left. {\left( {{{{x^6}} \over 6}\ln x - {{{x^6}} \over {36}}} \right)} \right|_1^2 = {{32} \over 3}\ln 2 - {7 \over 4}$$

b) Đặt

$$\left\{ \matrix{ u = x + 1 \hfill \cr dv = {e^x}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = dx \hfill \cr v = {e^x} \hfill \cr} \right.$$

$$\int\limits_0^1 {\left( {x + 1} \right)} {e^x}dx = \left. {\left( {x + 1} \right){e^x}} \right|_0^1 - \int\limits_0^1 {{e^x}dx = e}$$

c) Đặt $$I = \int\limits_0^\pi {{e^x}\cos xdx}$$

Đặt

$$\left\{ \matrix{ u = {e^x} \hfill \cr dv = \cos xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = {e^x}dx \hfill \cr v = {\mathop{\rm s}\nolimits} {\rm{inx}} \hfill \cr} \right.$$

Suy ra $$I = \left. {{e^x}{\mathop{\rm s}\nolimits} {\rm{inx}}} \right|_0^\pi - \int\limits_0^\pi {{e^x}\sin {\rm{x}}dx} = - \int\limits_0^\pi {{e^x}\sin {\rm{x}}dx}$$

Đặt

$$\left\{ \matrix{ u = {e^x} \hfill \cr dv = \sin {\rm{x}}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = {e^x}dx \hfill \cr v = - \cos x \hfill \cr} \right.$$

Do đó $$I = - \left[ {\left. {\left( { - {e^x}\cos x} \right)} \right|_0^\pi + \int\limits_0^\pi {{e^x}\cos xdx} } \right] = {e^\pi }\cos \pi - {e^0}.\cos 0 - I$$

$$\Rightarrow 2I = - {e^\pi } - 1 \Rightarrow I = - {1 \over 2}\left( {{e^\pi } + 1} \right)$$

b) Đặt

$$\left\{ \matrix{ u = x \hfill \cr dv = \cos xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = dx \hfill \cr v = {\mathop{\rm s}\nolimits} {\rm{inx}} \hfill \cr} \right.$$

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