Giải bài 6 trang 127 – SGK môn Giải tích lớp 12

một)

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\(\begin{aligned} \int\limits_{0}^{\frac{\pi }{2}}{\cos 2x{{\sin }^{2} }xdx}&=\dfrac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{\cos 2x\left( 1- cos 2x \right)dx} \\ & =\dfrac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{ cos 2xdx}-\dfrac{1}{4}\int\limits_{0}^{\frac{\pi }{2}}{\left( 1+\cos 4x \ Right)dx} \\ & =\left( \dfrac{\sin 2x-x}{4}-\dfrac{\sin 4x}{16} \right)\left| _{\begin{smalmatrix} \\ 0 \end{smalmatrix}}^{\frac{\pi }{2}} \right. \\ & =-\dfrac{ \pi }{8} \\ \end{alignment} \)

hai)

\(\begin{aligned} \int\limits_{-1}^{1}{\left| {{2}^{x}}-{{2}^{-x} } \right|dx} & =\int\limits_{-1}^{0}{\left( {{2}^{-x}}-{{2}^{x}} right)dx}+\int\limits_{0}^{1}{\left({{2}^{x}}-{{2}^{-x}}\right)dx} \\ & =-\left( \dfrac{{{2}^{x}}+{{2}^{-x}}}{\ln 2} \right)\left | _{\begin{smalmatrix} \\ -1 \end{smallmatrix}}^{\begin{smalmatrix} 0 \\ \end{smalmatrix}} \right.+\left ( \dfrac{{{2}^{x}}+{{2}^{-x}}}{\ln 2} \right)\left| _{\begin{smalmatrix} \ \ 0 \end{smalmatrix}}^{\begin{smalmatrix} 1 \\ \end{smallmatrix}} \right. \\ & =\dfrac{1}{\ ln 2}\left( -1-1+\dfrac{1}{2}+2+2+\dfrac{1}{2}-1-1 \right) \\ & = \dfrac{1}{\ln 2} \\ \end{alignment} \)

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c)

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\(\begin{aligned} \int\limits_{1}^{2}{\dfrac{\left( x+1 \right)\left( x+2 \ Phải)\left( x+3 \right)}{{{x}^{2}}}} &=\int\limits_{1}^{2}{\dfrac{{{x }^{3}}+6{{x}^{2}}+11x+6}{{{x}^{2}}}dx} \\ & =\int\limits_{1 }^{2}{\left( x+6+\dfrac{11}{x}+\dfrac{6}{{{x}^{2}}} \right)dx} \ & =\left( \dfrac{{{x}^{2}}}{2}+6x+11.\ln \left| x \right|-\dfrac{6}{ x} \right)\left| _{\begin{smallmatrix} \\ 1 \end{smalmatrix}}^{\begin{smalmatrix} 2 \\ \end{smalmatrix}} \Đúng.\\ &=2+12+11\ln 2-3-\dfrac{1}{2}-6+6 \\ & =\dfrac{21}{ 2}+11\ln 2 \\ \end{alignment} \)

d)

\(\begin{aligned} \int\limits_{0}^{2}{\dfrac{1}{{{x}^{2}}-2x-3}dx} &amp ; =\int\limits_{0}^{2}{\dfrac{1}{\left( x+1 \right)\left( x-3 \right)}dx} \ \ & =\dfrac{1}{4}\int\limits_{0}^{2}{\left( \dfrac{1}{x-3}-\dfrac{1} {x+1} \right)dx} \\ & =\dfrac{1}{4}\ln \left| \dfrac{x-3}{x+1} \right |\left| _{\begin{smallmatrix} \\ 0 \end{smalmatrix}}^{\begin{smalmatrix} 2 \\ \end{smalmatrix}} \right. \ & =\dfrac{1}{4}\left( \ln \dfrac{1}{3}-\ln 3 \right)=-\dfrac{\ln 3 }{2} \\ \end{alignment} \)

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e)

\( \begin{aligned} \int\limits_{0}^{\frac{\pi }{2}}{{{\left( \sin x+\cos x \right)} ^{2}}dx}&=\int\limits_{0}^{\frac{\pi }{2}}{\left( 1+\sin 2x right)dx} \\ & =\left( x-\dfrac{\cos 2x}{2} \right)\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{\pi }{2}} \right.\\ & =\dfrac{\pi }{2}+\dfrac{1 }{2}+\dfrac{1}{2} \\ & =\dfrac{\pi }{2}+1 \\ \end{aligned} \)

g)

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\(\begin{aligned} \int\limits_{0}^{\pi }{{{\left( x+\sin x \right)}^{2}}dx } &=\int\limits_{0}^{\pi }{\left( {{x}^{2}}+2x\sin x+{{\sin }^{2}} x \right)dx} \\ & =\int\limits_{0}^{\pi }{{{x}^{2}}dx}+2\int\limits_{ 0}^{\pi }{x\sin xdx}+\int\limits_{0}^{\pi }{{{\sin }^{2}}xdx} \\ &amp ; ={{i}_{1}}+2{{i}_{2}}+{{i}_{3}} \\ \end{aligned} \)

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\({{i}_{1}}=\dfrac{{{x}^{3}}}{3}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} \pi \\ \end{smallmatrix}} \right.=\dfrac{{{\pi }^{3}}} {3} \)

Will \(\left\{ \begin{aligned} & x=u \\ & \sin xdx=dv \\ \end{aligned} \right. \rightarrow \left\{ \begin{aligned} & du=dx \\ & v=-\cos x \\ \end{aligned} \right. \)

\(\begin{aligned} {{i}_{2}}&=-x\cos x\left| _{\begin{smalmatrix} \\ 0 \end {smallmatrix}}^{\begin{smallmatrix} \pi \\ \end{smallmatrix}} \right.+\int\limits_{{0}}^{{\pi}} {\cos xdx} \\ & =\pi +\sin x\left| _{\begin{smallmatrix} \\ 0 \end{smalmatrix}}^{\begin {smallmatrix} \pi \\ \end{smallmatrix}} \right.=\pi \\ \end{aligned} \)

\(\begin{aligned} {{i}_{3}}& =\int\limits_{0}^{\pi }{\dfrac{1-\cos 2x {2}dx} \\ & =\left( \dfrac{x}{2}-\dfrac{\sin 2x}{4} \right)\left| _{ begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} \pi \\ \end{smallmatrix}} \right. \\ & = dfrac{\pi }{2} \\ \end{alignment} \)

\(\rightarrow \int\limits_{0}^{\pi }{{{\left( x+\sin x \right)}^{2}}dx}= \dfrac{{{\pi }^{3}}}{3}+2\pi +\dfrac{\pi }{2}=\dfrac{{{\pi }^{3 } }}{3}+\dfrac{5\pi }{2}\)