Bài 2: Tích phân

ta có v=s'= \(3t^2-6t-9\)

gia tốc a=s"=\(6t-6\), thời điểm gia tốc triệt tiêu suy ra a=0 \(\Leftrightarrow\)t=1

\(\Rightarrow\)v= -12(m/s)

Đặt \(x=tant\) suy ra \(dx=\dfrac{dt}{cos^2t}\).
Đổi cận :
\(x=\sqrt{3}\Rightarrow t=arctan\sqrt{3}=\dfrac{\pi}{3}\).
\(x=1\Rightarrow t=arctan1=\dfrac{\pi}{4}\).
\(\int\limits^{\sqrt{3}}_1\dfrac{1}{\left(x^2+1\right)^2}dx\)\(=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\dfrac{dt}{\left(tan^2t+1\right)^2.cot^2t}=\)\(=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\dfrac{cos^4t}{cos^2t}dt=\)\(=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}cos^2tdt=\)\(=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\dfrac{1}{2}\left(1+cos2t\right)dt=\)\(=\dfrac{\pi}{24}+\dfrac{1}{4}\left(\dfrac{\sqrt{3}}{2}-1\right)\).

Câu a)

\(\int \frac{1}{\cos^4x}dx=\int \frac{\sin ^2x+\cos^2x}{\cos^4x}dx=\int \frac{\sin ^2x}{\cos^4x}dx+\int \frac{1}{\cos^2x}dx\)

Xét \(\int \frac{1}{\cos^2x}dx=\int d(\tan x)=\tan x+c\)

Xét \(\int \frac{\sin ^2x}{\cos^4x}dx=\int \frac{\tan ^2x}{\cos^2x}dx=\int \tan^2xd(\tan x)=\frac{\tan ^3x}{3}+c\)

Vậy :

\(\int \frac{1}{\cos ^4x}dx=\frac{\tan ^3x}{3}+\tan x+c\)

\(\Rightarrow \int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{dx}{\cos^4 x}=\)\(\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|\left ( \frac{\tan ^3 x}{3}+\tan x+c \right )=\frac{44}{9\sqrt{3}}\)

Câu b)

\(\int \frac{(x+1)^2}{x^2+1}dx=\int \frac{x^2+1+2x}{x^2+1}dx=\int dx+\int \frac{2xdx}{x^2+1}\)

\(=x+c+\int \frac{d(x^2+1)}{x^2+1}=x+\ln (x^2+1)+c\)

Do đó:

\(\int ^{1}_{0}\frac{(x+1)^2}{x^2+1}dx=\left.\begin{matrix} 1\\ 0\end{matrix}\right|(x+\ln (x^2+1)+c)=\ln 2+1\)

Câu c)

\(\int \frac{x^2+2\ln x}{x}dx=\int xdx+2\int \frac{2\ln x}{x}dx\)

\(=\frac{x^2}{2}+c+2\int \ln xd(\ln x)\)

\(=\frac{x^2}{2}+c+\ln ^2x\)

\(\Rightarrow \int ^{2}_{1}\frac{x^2+2\ln x}{x}dx=\left.\begin{matrix} 2\\ 1\end{matrix}\right|\left ( \frac{x^2}{2}+\ln ^2x +c \right )=\frac{3}{2}+\ln ^22\)

Câu d)

\(\int^{2}_{1} \frac{x^2+3x+1}{x^2+x}dx=\int ^{2}_{1}dx+\int ^{2}_{1}\frac{2x+1}{x^2+x}dx\)

\(=\left.\begin{matrix} 2\\ 1\end{matrix}\right|x+\int ^{2}_{1}\frac{d(x^2+x)}{x^2+x}=1+\left.\begin{matrix} 2\\ 1\end{matrix}\right|\ln |x^2+x|=1+\ln 6-\ln 2\)

\(=1+\ln 3\)

Câu e)

Xét \(\int 3x(x+\sqrt{x^2+16})dx=\int 3x^2dx+\int 3x\sqrt{x^2+16}dx\)

Có:

\(\int 3x^2dx=x^3+c\)

\(\int 3x\sqrt{x^2+16}dx=\frac{3}{2}\int \sqrt{x^2+16}d(x^2+16)\)

\(=\sqrt{(x^2+16)^3}+c\)

Do đó: \(\int 3x(x+\sqrt{x^2+16})dx=x^3+\sqrt{(x^2+16)^3}+c\)

\(\Rightarrow \int ^{3}_{0}3x(x+\sqrt{x^2+16})dx=\left.\begin{matrix} 3\\ 0\end{matrix}\right|(x^3+\sqrt{(x^2+16)^3}+c)=88\)

Lời giải:

Xét \(\int \frac{\tan ^2x-\cos ^2x}{\sin ^2x}dx=\int \frac{\tan ^2x}{\sin ^2x}dx-\int \frac{\cos ^2x}{\sin ^2x}dx\)

Có:

\(\int \frac{\tan ^2x}{\sin ^2x}dx=\int \frac{\sin ^2x}{\cos ^2x. \sin^2 x}dx=\int \frac{1}{\cos ^2x}dx\)

\(=\int d(\tan x)=\tan x+c\)

Và:

\(\int \frac{\cos ^2x}{\sin ^2x}dx=\int \frac{1-\sin ^2x}{\sin ^2x}dx=\int \frac{1}{\sin ^2x}dx-\int dx\)

\(=-\int d(\cot x)-x+c=-\cot x-x+c\)

Do đó:

\(\int \frac{\tan ^2x-\cos ^2x}{\sin ^2x}dx=\tan x+c-(-\cot x-x+c)=\tan x+\cot x+x+c\)

\(\Rightarrow \int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\tan ^2x-\cos ^2x}{\sin ^2x}dx=\frac{4\sqrt{3}}{3}+\frac{\pi}{3}-\frac{4\sqrt{3}}{3}-\frac{\pi}{6}=\frac{\pi}{6}\)

Lời giải:

Đặt \(x=\sqrt{3}\tan t(t\in (0; \frac{\pi}{2}))\)

\(\Rightarrow \sqrt{9+3x^2}=\sqrt{9+9\tan ^2t}=\sqrt{\frac{9}{\cos ^2t}}=\frac{3}{\cos t}\)

Khi đó \(I=\int \frac{3d(\sqrt{3}\tan t)}{3\cos t.\tan ^2t}=\int \frac{d(\sqrt{3}\tan t)}{\cos t.\tan ^2t}\)

\(=\int \frac{\sqrt{3}dt}{\cos ^3t\tan ^2t}=\sqrt{3}\int \frac{dt}{\cos ^3.\frac{\sin ^2t}{\cos ^2t}}\)

\(=\sqrt{3}\int \frac{dt}{\cos t\sin ^2t}\)

Đặt \(\left\{\begin{matrix} u=\frac{1}{\cos t}\\ dv=\frac{dt}{\sin ^2t}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{\sin t}{\cos ^2t}dt\\ v=-\cot t\end{matrix}\right.\)

Suy ra \(I=\sqrt{3}(\frac{-\cot t}{\cos t}+\int \frac{\cot t\sin t}{\cos ^2t}dt)\)

\(=\sqrt{3}(\frac{-\cot t}{\cos t}+\int \frac{dt}{\cos t})\)

\(=\sqrt{3}(\frac{-\cot t}{\cos t}+\int \frac{d(\sin t)}{1-\sin ^2t})\)

Phân tích:

\(\int \frac{d(\sin t)}{1-\sin ^2t}=\int \frac{dk}{1-k^2}=\frac{1}{2}\int \frac{dk}{1-k}+\frac{1}{2}\int \frac{dk}{1+k}=\frac{1}{2}\ln |k+1|-\frac{1}{2}\ln |1-k|+c\)

\(=\frac{1}{2}\ln |\frac{\sin t+1}{\sin t-1}|+c\)

Vậy \(I=\sqrt{3}(\frac{\cot t}{\cos t}+\frac{1}{2}\ln |\frac{\sin t+1}{\sin t-1}|)+c\)

Câu 1:

Ta có \(I_1=\int ^{1}_{0}\frac{4x+2}{x^2+x+1}dx=2\int ^{1}_{0}\frac{2x+1}{x^2+x+1}dx\)

\(=2\int ^{1}_{0}\frac{d(x^2+x+1)}{x^2+x+1}=2.\left.\begin{matrix} 1\\ 0\end{matrix}\right|\ln |x^2+x+1|=2\ln 3\)

Câu 2:

\(I_2=\int ^{1}_{0}\frac{4x+1}{(2-x)^4}dx=\int ^{1}_{0}\frac{4(x-2)+9}{(2-x)^4}dx\)

\(=4\int ^{1}_{0}\frac{dx}{(x-2)^3}+9\int \frac{dx}{(2-x)^4}=4\int ^{1}_{0}\frac{d(x-2)}{(x-2)^3}-9\int ^{1}_{0}\frac{d(2-x)}{(2-x)^4}\)

\(=4\int ^{-1}_{-2}\frac{dt}{t^3}-9\int ^{1}_{2}\frac{dk}{k^4}\) với \(x-2=t; 2-x=k\)

\(=4.\left.\begin{matrix} -1\\ -2\end{matrix}\right|\frac{t^{-3+1}}{-3+1}-9.\left.\begin{matrix} 1\\ 2\end{matrix}\right|\frac{k^{-4+1}}{-4+1}=\frac{9}{8}\)

Câu 3:

Phân số \(\frac{x^2+1}{(x^3+3x)^3}\) không xác định trên \([0;1]\); hàm không liên tục nên không có tích phân.

Ở tất cả các dạng bài như thế này em chỉ cần ghi nhớ công thức:

\(d(u(x))=u'(x)dx\)

Câu 1)

Ta có \(I_1=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} e^{\sin x}\cos xdx=\int _{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{\sin x}d(\sin x)\)

Đặt \(\sin x=t\Rightarrow I_1=\int ^{1}_{\frac{\sqrt{2}}{2}}e^tdt=\left.\begin{matrix} 1\\ \frac{\sqrt{2}}{2}\end{matrix}\right|e^t=e-e^{\frac{\sqrt{2}}{2}}\)

Câu 2)

\(I_2=\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}}e^{2\cos x+1}\sin xdx=\frac{-1}{2}\int ^\frac{\pi}{2}_{\frac{\pi}{4}}e^{2\cos x+1}d(2\cos x+1)\)

Đặt \(2\cos x+1=t\Rightarrow I_2=\frac{-1}{2}\int ^{1}_{1+\sqrt{2}}e^tdt\)

\(=\frac{-1}{2}.\left.\begin{matrix} 1\\ 1+\sqrt{2}\end{matrix}\right|e^t=\frac{-1}{2}(e-e^{1+\sqrt{2}})\)

Câu 3:

Có \(I_3=\int ^{e}_{1}\frac{e^{2\ln x+1}}{x}dx=\int ^{e}_{1}e^{2\ln x+1}d(\ln x)\)

\(=\frac{1}{2}\int ^{e}_{1}e^{2\ln x+1}d(2\ln x+1)\)

Đặt \(2\ln x+1=t\Rightarrow I_3=\frac{1}{2}\int ^{3}_{1}e^tdt=\frac{1}{2}.\left.\begin{matrix} 3\\ 1\end{matrix}\right|e^t=\frac{1}{2}(e^3-e)\)

Câu 4:

\(I_4=\int ^{1}_{0}xe^{x^2+2}dx=\frac{1}{2}\int ^{1}_{0}e^{x^2+2}d(x^2+2)\)

Đặt \(x^2+2=t\Rightarrow I_4=\frac{1}{2}\int ^{3}_{2}e^tdt=\frac{1}{2}.\left.\begin{matrix} 3\\ 2\end{matrix}\right|e^t=\frac{1}{2}(e^3-e^2)\)