Giải bài 3.6 trang 164 - SBT Đại số và Giải tích lớp 12

a) \(I_1=\int{x{{\left( 3-x \right)}^{5}}dx}\)

Đặt \(3-x=u\Rightarrow dx=-du\), ta có:

\(\begin{align} & {{I}_{1}}=\int{\left( u-3 \right){{u}^{5}}}du=\int{\left( {{u}^{6}}-3{{u}^{5}} \right)}du=\frac{1}{7}{{u}^{7}}-\frac{1}{2}{{u}^{6}}+C \\ & \,\,\,\,\,=\frac{1}{7}{{\left( 3-x \right)}^{7}}-\frac{1}{2}{{\left( 3-x \right)}^{6}}+C \\ \end{align} \)

b) Ta có:

 \(\int{{{\left( {{2}^{x}}-{{3}^{x}} \right)}^{2}}dx}=\int{{{\left( {{2}^{2x}}-2.2^x{{3}^{x}}+3^{2x} \right)}}dx}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int 4^xdx-2\int 6^xdx+\int 9^x dx\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{4^x}{ln4}-2\dfrac{6^x}{ln6}+\dfrac{9^x}{ln9}+C\)

c) \(I_2= \int{x\sqrt{2-5x}dx}\)

Đặt \(\sqrt{2-5x}=u\Rightarrow 2-5x=u^2\Rightarrow dx=-\dfrac{2}{5}udu\)

Ta có: \(x=\dfrac{2-u}{5}\) nên

\(\begin{align} & {{I}_{2}}=-\int{\frac{2-u}{5}.u.\frac{2}{5}udu} \\ & \,\,\,\,\,\,=\int{\left( -\frac{4}{25}{{u}^{2}}+\frac{2}{25}{{u}^{3}} \right)}du \\ & \,\,\,\,\,\,=-\frac{4}{75}{{u}^{3}}+\frac{1}{50}{{u}^{4}}+C \\ & \,\,\,\,\,=-\frac{4}{75}{{\left( 2-5x \right)}^{\frac{3}{2}}}+\frac{1}{50}{{\left( 2-5x \right)}^{2}}+C \\ \end{align} \)

c) \(I_3=\int{\dfrac{\ln \left( \cos x \right)}{{{\cos }^{2}}x}dx}\)

Đặt \(\left\{ \begin{aligned} & u=\ln \left( \cos x \right) \\ & dv=\frac{1}{{{\cos }^{2}}x}dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=-\frac{\sin x}{\cos x}dx \\ & v=\tan x \\ \end{aligned} \right. \)

Ta có: 

\(\begin{align} & {{I}_{3}}=\tan x\ln \left( \cos x \right)+\int{\tan x.\frac{\sin x}{\cos x}dx} \\ & \,\,\,\,\,\,=\tan x\ln \left( \cos x \right)+\int{\dfrac{\sin^2x}{\cos^2x}dx} \\ & \,\,\,\,\,\,=\tan x\ln \left( \cos x \right)+\int{\dfrac{1-\cos^2x}{\cos^2x}dx} \\ & \,\,\,\,\,\,=\tan x\ln \left( \cos x \right)+\int{\left(\dfrac{1}{\cos^2x}-1\right)dx} \\ & \,\,\,\,\,\,=\tan x\ln \left( \cos x \right)+\tan x-x+C \end{align} \)

e) \(I_4=\int{\dfrac{x}{{{\sin }^{2}}x}dx}\)

Đặt \(\left\{ \begin{aligned} & u=x \\ & dv=\frac{1}{{{\sin }^{2}}x}dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v=-\cot x \\ \end{aligned} \right. \)

Ta có: 

\(\begin{align} & {{I}_{4}}=-x\cot x+\int{\cot xdx} \\ & \,\,\,\,\,=-x\cot x+\int{\frac{\cos x}{\sin x}dx} \\ \end{align}\ \)

Đặt \(t=\sin x\Rightarrow dt=\cos xdx\). Suy ra:

\(\begin{align} & {{I}_{4}}=-x\cot x+\int{\frac{dt}{t}} \\ & \,\,\,\,\,=-x\cot x+\ln \left| t \right|+C \\ \\ & \,\,\,\,\,=-x\cot x+\ln \left| \operatorname{s}\text{inx} \right|+C \\ \end{align} \)

g) \(I_5=\int{\dfrac{x+1}{\left( x-2 \right)\left( x+3 \right)}dx}\)

Ta có: \(\dfrac{x+1}{\left( x-2 \right)\left( x+3 \right)}=\dfrac{3}{5(x-2)}+\dfrac{2}{5(x+3)}\)

Khi đó:

 \(\begin{align} & {{I}_{5}}=\int{\frac{3}{5\left( x-2 \right)}dx}+\int{\frac{2}{5\left( x+3 \right)}dx} \\ & \,\,\,\,\,=\frac{3}{5}\ln \left| x-2 \right|+\frac{2}{5}\ln \left| x+3 \right|+C \\ \end{align} \)\

h) \(I_6=\int{\dfrac{1}{1-\sqrt{x}}dx}\)

Đặt \(u=\sqrt{x}\Rightarrow u^2=x\Rightarrow dx=2udu \)

Ta có:

\(\begin{align} & {{I}_{6}}=\int{\frac{2udu}{1-u}}=\int{\left( -2+\frac{2}{1-u} \right)du} \\ & \,\,\,\,\,=-2u+2\ln \left| 1-u \right|+C \\ \\ & \,\,\,\,\,=-2\sqrt{x}+2ln\left| 1-\sqrt{x} \right|+C \\ \end{align} \)

i) \( \int{\sin 3x\cos 2xdx}=\dfrac{1}{2} \int{(\sin x+\sin 5x)dx}=-\dfrac{1}{2}\left( \cos x+\dfrac{1}{5} \cos 5x\right)+C\)

Phương pháp đổi biến số

Nếu \(\int{f\left( u \right)}du=F\left( u \right)+C\) và \(u=u\left( x \right)\) là hàm số có đạo hàm liên tục thì

\(\int{f\left( u\left( x \right) \right)u'\left( x \right)dx=F\left( u\left( x \right) \right)+C}\)

Phương pháp tính nguyên hàm từng phần: Nếu hai hàm số u=u(x) và v=v(x) có đạo hàm liên tục trên K thì

\(\int{u\left( x \right)v'\left( x \right)dx=u\left( x \right)v\left( x \right)-\int{u'\left( x \right)v\left( x \right)dx}}\)

Hay \(\int{udv=uv-\int{vdu}}\)