a) Đặt
\(\left\{ \matrix{
u = {x^2} \hfill \cr
dv = \cos 2xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = 2xdx \hfill \cr
v = {1 \over 2}\sin 2x \hfill \cr} \right.\)
Do đó \(\int {{x^2}\cos 2xdx = {1 \over 2}{x^2}\sin 2x} - \int {x\sin 2xdx\,\,\,\left( 1 \right)} \)
Tính \(\int {x\sin 2xdx} \)
Đặt
\(\left\{ \matrix{ u = x \hfill \cr dv = \sin 2xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = dx \hfill \cr v = - {1 \over 2}\cos 2x \hfill \cr} \right.\)
\( \Rightarrow \int {x\sin 2xdx = - {1 \over 2}x\cos 2x + {1 \over 2}\int {\cos 2xdx = - {1 \over 2}x\cos 2x - {1 \over 4}\sin 2x + C} } \)
Thay vào (1) ta được \(\int {{x^2}\cos 2xdx = {1 \over 2}{x^2}\sin 2x + {1 \over 2}x\cos 2x + {1 \over 4}\sin 2x + C} \)
b) Đặt
\(\left\{ \matrix{ u = \ln x \hfill \cr dv = \sqrt x dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = {{dx} \over x} \hfill \cr v = {2 \over 3}{x^{{3 \over 2}}} \hfill \cr} \right.\)
\( \Rightarrow \int {\sqrt x } \ln xdx = {2 \over 3}{x^{{3 \over 2}}}\ln x - {2 \over 3}\int {{x^{{1 \over 2}}}dx} \)
\( = {2 \over 3}{x^{{3 \over 2}}}\ln x - {2 \over 3}.{2 \over 3}{x^{{3 \over 2}}} + C = {2 \over 3}\sqrt {{x^3}} \ln x - {4 \over 9}\sqrt {{x^3}} + C\)
c) Đặt \(u = {\mathop{\rm s}\nolimits} {\rm{inx}} \Rightarrow du = \cos xdx\)
\( \Rightarrow \int {{{\sin }^4}x\cos xdx = } \int {{u^4}du = {{{u^5}} \over 5} + C = {1 \over 5}{{\sin }^5}x} + C.\)
d) Đặt \(u = {x^2} \Rightarrow du = 2xdx \Rightarrow xdx = {1 \over 2}du\)
\( \Rightarrow \int {x\cos \left( {{x^2}} \right)dx = {1 \over 2}\int {\cos udu = {1 \over 2}\sin u + C = {1 \over 2}{\mathop{\rm s}\nolimits} {\rm{in}}{{\rm{x}}^2} + C.} } \)
