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lâu ko làm tích phân cũng quên béng đi rồi những câu này cũng không khó chú ý 1 chút là làm đc ak ,
trong cái căn bậc 2 nhé 3+2x-x^2= -((x-1)^2+2)) sau do dat x-1=a nen x+1=a+2 thay vap bieu tu lam binh thuong la ra thoi ak
\(I_1=3\int_1^2x^2dx+\int_1^2\cos xdx+\int_1^2\frac{dx}{x}=x^3\)\(|^2 _1\)+\(\sin x\)\(|^2_1\) +\(\ln\left|x\right|\)\(|^2_1\)
\(=\left(8-1\right)+\left(\sin2-\sin1\right)+\left(\ln2-\ln1\right)\)
\(=7+\sin2-\sin1+\ln2\)
b) \(I_2=4\int_1^2\frac{dx}{x}-5\int_1^2x^4dx+2\int_1^2\sqrt{x}dx\)
\(=4\left(\ln2-\ln1\right)-\left(2^5-1^5\right)+\frac{4}{3}\left(2\sqrt{2}-1\sqrt{1}\right)\)
\(=4\ln2+\frac{8\sqrt{2}}{3}-32\frac{1}{3}\)
\(I_1=\int\limits^2_0f\left(2x\right)dx\)
Đặt \(2x=t\Rightarrow dx=\frac{dt}{2}\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=2\Rightarrow t=4\end{matrix}\right.\)
\(\Rightarrow I_1=\int\limits^4_0f\left(t\right).\frac{dt}{2}=\frac{1}{2}\int\limits^4_0f\left(t\right)dt=\frac{1}{2}\int\limits^4_0f\left(x\right)dx=\frac{1}{2}.2018=1009\)
\(I_2=\int\limits^2_{-2}f\left(2-x\right)dx\)
Đặt \(2-x=t\Rightarrow dx=-dt\); \(\left\{{}\begin{matrix}x=-2\Rightarrow t=4\\x=2\Rightarrow t=0\end{matrix}\right.\)
\(\Rightarrow I_2=\int\limits^0_4f\left(t\right).\left(-dt\right)=\int\limits^4_0f\left(t\right)dt=\int\limits^4_0f\left(x\right)dx=2018\)
\(\Rightarrow I=I_1+I_2=1009+2018=3027\)
Ko thể dịch nổi đề câu 1 a;b, chỉ đoán thôi. Còn câu 2 thì thực sự là chẳng hiểu bạn viết cái gì nữa? Chưa bao giờ thấy kí hiệu tích phân đi kèm kiểu đó
Câu 1:
a/ \(\int\frac{2x+4}{x^2+4x-5}dx=\int\frac{d\left(x^2+4x-5\right)}{x^2+4x-5}=ln\left|x^2+4x-5\right|+C\)
b/ \(\int\frac{1}{x.lnx}dx\)
Đặt \(t=lnx\Rightarrow dt=\frac{dx}{x}\)
\(\Rightarrow I=\int\frac{dt}{t}=ln\left|t\right|+C=ln\left|lnx\right|+C\)
c/ \(I=\int x.sin\frac{x}{2}dx\)
Đặt \(\left\{{}\begin{matrix}u=x\\dv=sin\frac{x}{2}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-2cos\frac{x}{2}\end{matrix}\right.\)
\(\Rightarrow I=-2x.cos\frac{x}{2}+2\int cos\frac{x}{2}dx=-2x.cos\frac{x}{2}+4sin\frac{x}{2}+C\)
d/ Đặt \(\left\{{}\begin{matrix}u=ln\left(2x\right)\\dv=x^3dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{2dx}{2x}=\frac{dx}{x}\\v=\frac{1}{4}x^4\end{matrix}\right.\)
\(\Rightarrow I=\frac{1}{4}x^4.ln\left(2x\right)-\frac{1}{4}\int x^3dx=\frac{1}{4}x^4.ln\left(2x\right)-\frac{1}{16}x^4+C\)
Đặt \(3-2x=t\Rightarrow dx=-\frac{1}{2}dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=3\\x=2\Rightarrow t=-1\end{matrix}\right.\)
\(\Rightarrow P=\int\limits^{-1}_3\left[f\left(t\right)+2019\right].\left(-\frac{1}{2}\right)dt=\frac{1}{2}\int\limits^3_{-1}f\left(t\right)dt+\int\limits^3_{-1}\frac{2019}{2}dt\)
\(=\frac{15}{2}+\frac{2019}{2}.4=\frac{8091}{2}\)
1)
Ta có:
\(\int (2-\cot ^2x)dx=\int (2-\frac{\cos ^2x}{\sin ^2x})dx\)
\(=\int (2-\frac{1-\sin ^2x}{\sin ^2x})dx=\int (3-\frac{1}{\sin ^2x})dx=3\int dx-\int \frac{dx}{\sin ^2x}\)
\(=3x+\int d(\cot x)=3x+\cot x+c\)
\(\Rightarrow \int ^{\frac{\pi}{2}}_{\frac{\pi}{3}}(2-\cot ^2x)dx=\left.\begin{matrix} \frac{\pi}{2}\\ \frac{\pi}{3}\end{matrix}\right|(3x+\cot x+c)=\frac{\pi}{2}-\frac{\sqrt{3}}{3}\)
3)
Xét \(\int (2\tan x-3\cot x)^2dx\)
\(=\int (4\tan ^2x+9\cot ^2x-12)dx\)
\(=\int (\frac{4\sin ^2x}{\cos ^2x}+\frac{9\cos ^2x}{\sin ^2x}-12)dx\)
\(=\int (\frac{4(1-\cos ^2x)}{\cos ^2x}+\frac{9(1-\sin ^2x)}{\sin ^2x}-12)dx\)
\(=\int (\frac{4}{\cos ^2x}+\frac{9}{\sin ^2x}-25)dx\)
\(=4\int d(\tan x)-9\int d(\cot x)-25\int dx\)
\(=4\tan x-9\cot x-25x+c\)
Do đó:
\(\int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}(2\tan x-3\cot x)^2dx=\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|(4\tan x-9\cot x-25x+c)=\frac{26\sqrt{3}}{3}-\frac{25\pi}{6}\)
2)
Xét \(\int (\tan x+\cot x)^2dx=\int (\tan ^2x+\cot ^2x+2)dx\)
\(=\int (\frac{\sin ^2x}{\cos^2 x}+\frac{\cos ^2x}{\sin ^2x}+2)dx\)
\(=\int (\frac{1-\cos ^2x}{\cos ^2x}+\frac{1-\sin ^2x}{\sin ^2x}+2)dx\)
\(=\int (\frac{1}{\cos ^2x}+\frac{1}{\sin ^2x})dx\)
\(=\int d(\tan x)-\int d(\cot x)=\tan x-\cot x+c\)
Do đó:
\(\int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}(\tan x+\cot x)^2dx=\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|(\tan x-\cot x+c)=2\sqrt{3}-\frac{2\sqrt{3}}{3}\)
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)
\(\int^{2}_{1} \frac{x^2+3x+1}{x^2+x}dx=\int ^{2}_{1}dx+\int ^{2}_{1}\frac{2x+1}{x^2+x}dx\)
\(=\left.\begin{matrix} 2\\ 1\end{matrix}\right|x+\int ^{2}_{1}\frac{d(x^2+x)}{x^2+x}=1+\left.\begin{matrix} 2\\ 1\end{matrix}\right|\ln |x^2+x|=1+\ln 6-\ln 2\)
\(=1+\ln 3\)