a) Đặt
\(\left\{ \matrix{ u = x \hfill \cr dv = \sin {x \over 2}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = dx \hfill \cr v = - 2\cos {x \over 2} \hfill \cr} \right.\)
Do đó \(\int {x\sin x{x \over 2}dx} = - 2x\cos {x \over 2} + 2\int {\cos {x \over 2}dx = - 2x\cos {x \over 2} + 4\sin {x \over 2} + C} \)
b) Đặt
\(\left\{ \matrix{ u = {x^2} \hfill \cr dv = \cos xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = 2xdx \hfill \cr v = {\mathop{\rm s}\nolimits} {\rm{inx}} \hfill \cr} \right.\)
Do đó \(\int {{x^2}} \cos xdx = {x^2}{\mathop{\rm s}\nolimits} {\rm{inx}} - 2\int {x\sin xdx\,\,\,\,\,\,\,\left( 1 \right)} \)
Tính \(\int {x\sin xdx} \)
Đặt
\(\left\{ \matrix{ u = x \hfill \cr dv = \sin {\rm{x}}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = dx \hfill \cr v = - \cos x \hfill \cr} \right.\)
\( \Rightarrow \int {x\sin xdx = - x\cos x + \int {\cos xdx = - x\cos x + {\mathop{\rm s}\nolimits} {\rm{inx}} + } } \,C\)
Thay vào (1) ta được: \(\int {{x^2}\cos xdx = {x^2}{\mathop{\rm s}\nolimits} {\rm{inx}} + 2x\cos x - 2\sin x + C} \)
c) Đặt
\(\left\{ \matrix{ u = x \hfill \cr dv = {e^x}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = dx \hfill \cr v = {e^x} \hfill \cr} \right.\)
Do đó \(\int {x{e^x}dx = x{e^x} - \int {{e^x}dx} = x{e^x} - {e^x}} + C\)
d) Đặt
\(\left\{ \matrix{ u = \ln x \hfill \cr dv = {x^3}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = {1 \over x}dx \hfill \cr v = {{{x^4}} \over 4} \hfill \cr} \right.\)
Do đó \(\int {{x^3}\ln xdx = {1 \over 4}{x^4}\ln x} - {1 \over 4}\int {{x^3}dx} = {1 \over 4}x^4\ln x - {{{x^4}} \over {16}} + C\)
