Bài tập 18 trang 161 SGK Toán 12 NC

Bài tập 18 trang 161 SGK Toán 12 NC

a) Đặt \(\left\{ \begin{array}{l}
u = lnx\\
dv = {x^5}dx
\end{array} \right. \)

\(\Rightarrow \left\{ \begin{array}{l}
du = \frac{{dx}}{x}\\
v = \frac{{{x^6}}}{6}
\end{array} \right.\)

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

b) Đặt 

\(\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{u = x + 1}\\
{dv = {e^x}dx}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = dx}\\
{v = {e^x}}
\end{array}} \right.}\\
\begin{array}{l}
\int\limits_0^1 {\left( {x + 1} \right)} {e^x}dx\\
 = \left. {\left( {x + 1} \right){e^x}} \right|_0^1 - \int\limits_0^1 {{e^x}} dx = e
\end{array}
\end{array}\)

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

Đặt 

\(\begin{array}{l}
\left\{ {\begin{array}{*{20}{l}}
{u = {e^x}}\\
{dv = \cos xdx}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = {e^x}dx}\\
{v = \sin x}
\end{array}} \right.\\
 \Rightarrow I = \left. {{e^x}\sin x} \right|_0^\pi  - \int\limits_0^\pi  {{e^x}} {\rm{sinx}}dx\\
 =  - \int\limits_0^\pi  {{e^x}} {\rm{sinx}}dx
\end{array}\)

Đặt \(\left\{ \begin{array}{l}
u = {e^x}\\
dv = sinxdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = {e^x}dx\\
v =  - \cos x
\end{array} \right.\)

Do đó: 

\(\begin{array}{*{20}{l}}
\begin{array}{l}
I =  - \left[ {( - {e^x}cosx)|_0^\pi  + \int\limits_0^\pi  {{e^x}} \cos xdx} \right]\\
 = {e^\pi }\cos \pi  - {e^0}.\cos 0 - I
\end{array}\\
{ \Rightarrow 2I =  - {e^\pi } - 1 \Rightarrow I =  - \frac{1}{2}\left( {{e^\pi } + 1} \right)}
\end{array}\)

d) Đặt \(\left\{ \begin{array}{l}
u = x\\
dv = \cos xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = dx\\
v = \sin x
\end{array} \right.\)

Do đó:

\(\begin{array}{l}
\int\limits_0^{\frac{\pi }{2}} x \cos xdx = x\sin x|_0^{\frac{\pi }{2}} - \int\limits_0^{\frac{\pi }{2}} {\sin x} dx\\
 = (x\sin x + \cos x)|_0^{\frac{\pi }{2}} = \frac{\pi }{2} - 1
\end{array}\)

-- Mod Toán 12