Tìm nguyên hàm của các hàm số sau:
LG a
\(f\left( x \right) = {x^2}\cos 2x;\)
Lời giải chi tiết:
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
\(\left\{ \matrix{
u = x \hfill \cr
dv = \sin 2xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = dx \hfill \cr
v = - {1 \over 2}\cos 2x \hfill \cr} \right.\)
\( \Rightarrow \int {x\sin 2xdx }\) \(= - {1 \over 2}x\cos 2x + {1 \over 2}\int {\cos 2xdx }\) \( = - \dfrac{1}{2}x\cos 2x + \dfrac{1}{2}.\dfrac{{ - \cos 2x}}{2} + {C_1}\) \( = - {1 \over 2}x\cos 2x - {1 \over 4}\sin 2x + C_1 \)
Thay vào (1) ta được \(\int {{x^2}\cos 2xdx }\)
\( = \dfrac{1}{2}{x^2}\sin 2x \) \(- \left( { - \dfrac{1}{2}x\cos 2x - \dfrac{1}{4}\sin 2x + {C_1}} \right)\)
\(= {1 \over 2}{x^2}\sin 2x + {1 \over 2}x\cos 2x + {1 \over 4}\sin 2x + C \)
LG b
\(f\left( x \right) = \sqrt x \ln x;\)
Lời giải chi tiết:
Đặt
\(\left\{ \matrix{
u = \ln x \hfill \cr
dv = \sqrt x dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = {{dx} \over x} \hfill \cr
v = {2 \over 3}{x^{{3 \over 2}}} \hfill \cr} \right.\)
\( \Rightarrow \int {\sqrt x \ln xdx} \)\( = \dfrac{2}{3}{x^{\dfrac{3}{2}}}\ln x - \int {\dfrac{2}{3}{x^{\dfrac{3}{2}}}.\dfrac{1}{x}dx} \) \( = \dfrac{2}{3}{x^{\dfrac{3}{2}}}\ln x - \dfrac{2}{3}\int {{x^{\dfrac{1}{2}}}dx} \) \( = \dfrac{2}{3}{x^{\dfrac{3}{2}}}\ln x - \dfrac{2}{3}.\dfrac{{{x^{\dfrac{1}{2} + 1}}}}{{\dfrac{1}{2} + 1}} + C\) \( = \dfrac{2}{3}{x^{\dfrac{3}{2}}}\ln x - \dfrac{2}{3}.\dfrac{{{x^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}} + C\) \( = \dfrac{2}{3}{x^{\dfrac{3}{2}}}\ln x - \dfrac{2}{3}.\dfrac{2}{3}{x^{\dfrac{3}{2}}} + C\) \( = \dfrac{2}{3}x\sqrt x \ln x - \dfrac{4}{9}x\sqrt x + C\)
LG c
\(f\left( x \right) = {\sin ^4}x\cos x;\)
Lời giải chi tiết:
Đặt \(u = {\mathop{\rm s}\nolimits} {\rm{inx}} \Rightarrow du = \cos xdx\)
\(\int {{{\sin }^4}x\cos xdx} = \int {{u^4}du} \) \( = \dfrac{{{u^5}}}{5} + C = \dfrac{{{{\sin }^5}x}}{5} + C\)
LG d
\(f\left( x \right) = x\cos \left( {{x^2}} \right);\)
Lời giải chi tiết:
Đặt \(u = {x^2} \Rightarrow du = 2xdx \Rightarrow xdx = {1 \over 2}du\)
\( \Rightarrow \int {x\cos \left( {{x^2}} \right)dx}\) \( = {1 \over 2}\int {\cos udu}\) \( = {1 \over 2}\sin u + C \) \(= {1 \over 2}\sin (x^2) + C. \)
