\(\int_0^{\frac{\Pi}{2}}c\text{os}^2x\left(1-sin^3x\right)dx\)
2) \(\int_0^{\frac{\Pi}{4}}\frac{sin\left(x-\frac{\Pi}{4}\right)}{sin2x+2\left(1+s\text{inx}+c\text{ox}\right)}dx\)
hộ mk vs nha
Những câu hỏi liên quan
1)intsqrt{frac{1-sqrt{x}}{1+sqrt{x}}}dx
2)intfrac{dx}{left(e^x+1right)left(x^2+1right)}
3)intfrac{1+2xsqrt{1-x^2}+2x^2}{1+x+sqrt{1+x^2}}dx
4)intfrac{sin^6x+ctext{os}^6x}{1+6^x}dx
5)int_0^{frac{pi}{2}}frac{sqrt{ctext{os}x}}{sqrt{stext{inx}}+sqrt{ctext{os}x}}dx
6)intfrac{x^4}{2^x+1}dx
7)int_0^{frac{pi^2}{4}}sinsqrt{x}dx
8)intsqrt[6]{1-ctext{os}^3x}.stext{inx}.ctext{os}^5xdx
9)intsqrt{frac{1}{4x}+frac{sqrt{x}+e^x}{sqrt{x}.e^x}}dx
10)intfrac{ctext{os}x+stext{inx}}{left(e^xstext{inx}+1right)...
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1)\(\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx\)
2)\(\int\frac{dx}{\left(e^x+1\right)\left(x^2+1\right)}\)
3)\(\int\frac{1+2x\sqrt{1-x^2}+2x^2}{1+x+\sqrt{1+x^2}}\)dx
4)\(\int\frac{sin^6x+c\text{os}^6x}{1+6^x}dx\)
5)\(\int_0^{\frac{\pi}{2}}\frac{\sqrt{c\text{os}x}}{\sqrt{s\text{inx}}+\sqrt{c\text{os}x}}dx\)
6)\(\int\frac{x^4}{2^x+1}dx\)
7)\(\int_0^{\frac{\pi^2}{4}}sin\sqrt{x}dx\)
8)\(\int\sqrt[6]{1-c\text{os}^3x}.s\text{inx}.c\text{os}^5xdx\)
9)\(\int\sqrt{\frac{1}{4x}+\frac{\sqrt{x}+e^x}{\sqrt{x}.e^x}}dx\)
10)\(\int\frac{c\text{os}x+s\text{inx}}{\left(e^xs\text{inx}+1\right)s\text{inx}}dx\)
cho hàm số y=f(x) liên tục trên [0;π/2] thỏa \(\int_0^{\frac{\pi}{2}}f^2\left(x\right)dx=3\pi\) , \(\int_0^{\pi}\left(\sin x-x\right)f'\left(\frac{x}{2}\right)dx=6\pi\) ; \(f\left(\frac{\pi}{2}\right)=0\) Tính \(\int_0^{\frac{\pi}{2}}\left(f''\left(x\right)\right)^3dx\)
giúp em với ạ.
\(\int\frac{tan^3x}{c\text{os}2x}dx\)
2) \(\int\frac{xe^x\left(4+4\left(s\text{inx}+c\text{os}x\right)+sin2x\right)}{\left(1+c\text{os}x\right)^2}\)
1)
\(\int\frac{tan^3x}{cos2x}dx=\int\frac{sin^3x}{cos^3x\cdot\left(2cos^2x-1\right)}dx=\int\frac{1-cos^2x}{cos^3x\left(2cos^2x-1\right)}\cdot sinx\cdot dx\\ =\int\frac{1-cos^2x}{cos^3x\left(2cos^2x-1\right)}d\left(cosx\right)=...\)
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1.\(\int_0^1x\left(e^{2x^{ }}-\frac{x^2}{\sqrt{4-x^2}}\right)dx\)
2.Giải pt \(\sin\left(2x+\frac{\pi}{2}\right)+\cos x-\sin x=0\)
Tính tích phân các hàm lượng giác sau :a) I_1int_1^2left(3x^2+cos x+frac{1}{x}right)dxb) I_2int_1^2left(frac{4}{x}-5x^2+2sqrt{x}right)dxc) I_3int_a^bfrac{left|xright|}{x}dx, với ab0d) I_5int_0^{frac{pi}{2a}}left(x+3right)sin ax.dx với a0e)I_4int_0^{pi}sqrt{frac{1+cos2x}{2}}dx
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Tính tích phân các hàm lượng giác sau :
a) \(I_1=\int_1^2\left(3x^2+\cos x+\frac{1}{x}\right)dx\)
b) \(I_2=\int_1^2\left(\frac{4}{x}-5x^2+2\sqrt{x}\right)dx\)
c) \(I_3=\int_a^b\frac{\left|x\right|}{x}dx\), với ab>0
d) \(I_5=\int_0^{\frac{\pi}{2a}}\left(x+3\right)\sin ax.dx\) với a>0
e)\(I_4=\int_0^{\pi}\sqrt{\frac{1+\cos2x}{2}}dx\)
\(I_1=3\int_1^2x^2dx+\int_1^2\cos xdx+\int_1^2\frac{dx}{x}=x^3\)\(|^2 _1\)+\(\sin x\)\(|^2_1\) +\(\ln\left|x\right|\)\(|^2_1\)
\(=\left(8-1\right)+\left(\sin2-\sin1\right)+\left(\ln2-\ln1\right)\)
\(=7+\sin2-\sin1+\ln2\)
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b) \(I_2=4\int_1^2\frac{dx}{x}-5\int_1^2x^4dx+2\int_1^2\sqrt{x}dx\)
\(=4\left(\ln2-\ln1\right)-\left(2^5-1^5\right)+\frac{4}{3}\left(2\sqrt{2}-1\sqrt{1}\right)\)
\(=4\ln2+\frac{8\sqrt{2}}{3}-32\frac{1}{3}\)
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c) Ta cần xét 2 trường hợp 1) 0<a<b và 2) a<b<0
1) Nếu 0<a<b, khi đó \(f\left(x\right)=\frac{\left|x\right|}{x}=1\) vì \(x>0\)
Do đó
\(\int_a^bf\left(x\right)dx=\int_a^bdx=b-a\)
2) Nếu a<b<0, khi đó \(f\left(x\right)=\frac{\left|x\right|}{x}=\frac{-x}{x}=1\) vì \(x<0\)
Do đó :
\(\int_a^bf\left(x\right)dx=\int_a^b\left(-1\right)dx=-\left(b-a\right)=a-b\)
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Xem thêm câu trả lời
1) int lnfrac{left(1+stext{inx}right)^{1+ctext{os}x}}{1+ctext{os}x}dx
2) intleft(xlnxright)^2dx
3) intfrac{3xcosx+2}{1+cot^2x}dx
4)intfrac{2}{ctext{os}2x-7}dx
5)intfrac{1+xleft(2lnx-1right)}{xleft(x+1right)^2}dx
6) intfrac{1-x^2}{left(1+x^2right)^2}dx
7)int e^xfrac{1+stext{inx}}{1+ctext{os}x}dx
8) int lnleft(frac{x+1}{x-1}right)dx
9)intfrac{xlnleft(1+xright)}{left(1+x^2right)^2}dx
10) intfrac{lnleft(x-1right)}{left(x-1right)^4}dx
11)intfrac{x^3lnx}{sqrt{x^2+1}}dx
12)intfrac{xe^x}{_{ }...
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1) \(\int ln\frac{\left(1+s\text{inx}\right)^{1+c\text{os}x}}{1+c\text{os}x}dx\)
2) \(\int\left(xlnx\right)^2dx\)
3) \(\int\frac{3xcosx+2}{1+cot^2x}dx\)
4)\(\int\frac{2}{c\text{os}2x-7}dx\)
5)\(\int\frac{1+x\left(2lnx-1\right)}{x\left(x+1\right)^2}dx\)
6) \(\int\frac{1-x^2}{\left(1+x^2\right)^2}dx\)
7)\(\int e^x\frac{1+s\text{inx}}{1+c\text{os}x}dx\)
8) \(\int ln\left(\frac{x+1}{x-1}\right)dx\)
9)\(\int\frac{xln\left(1+x\right)}{\left(1+x^2\right)^2}dx\)
10) \(\int\frac{ln\left(x-1\right)}{\left(x-1\right)^4}dx\)
11)\(\int\frac{x^3lnx}{\sqrt{x^2+1}}dx\)
12)\(\int\frac{xe^x}{_{ }\left(e^x+1\right)^2}dx\)
13) \(\int\frac{xln\left(x+\sqrt{1+x^2}\right)}{x+\sqrt{1+x^2}}dx\)
giúp mk đc con nào thì giúp nha
Câu 2)
Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2\frac{\ln x}{x}dx\\ v=\frac{x^3}{3}\end{matrix}\right.\Rightarrow I=\frac{x^3}{3}\ln ^2x-\frac{2}{3}\int x^2\ln xdx\)
Đặt \(\left\{\begin{matrix} k=\ln x\\ dt=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dk=\frac{dx}{x}\\ t=\frac{x^3}{3}\end{matrix}\right.\Rightarrow \int x^2\ln xdx=\frac{x^3\ln x}{3}-\int \frac{x^2}{3}dx=\frac{x^3\ln x}{3}-\frac{x^3}{9}+c\)
Do đó \(I=\frac{x^3\ln^2x}{3}-\frac{2}{9}x^3\ln x+\frac{2}{27}x^3+c\)
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Câu 3:
\(I=\int\frac{2}{\cos 2x-7}dx=-\int\frac{2}{2\sin^2x+6}dx=-\int\frac{dx}{\sin^2x+3}\)
Đặt \(t=\tan\frac{x}{2}\Rightarrow \left\{\begin{matrix} \sin x=\frac{2t}{t^2+1}\\ dx=\frac{2dt}{t^2+1}\end{matrix}\right.\)
\(\Rightarrow I=-\int \frac{2dt}{(t^2+1)\left ( \frac{4t^2}{(t^2+1)^2}+3 \right )}=-\int\frac{2(t^2+1)dt}{3t^4+10t^2+3}=-\int \frac{2d\left ( t-\frac{1}{t} \right )}{3\left ( t-\frac{1}{t} \right )^2+16}=\int\frac{2dk}{3k^2+16}\)
Đặt \(k=\frac{4}{\sqrt{3}}\tan v\). Đến đây dễ dàng suy ra \(I=\frac{-1}{2\sqrt{3}}v+c\)
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Câu 6)
\(I=-\int \frac{\left ( 1-\frac{1}{x^2} \right )dx}{x^2+2+\frac{1}{x^2}}=-\int \frac{d\left ( x+\frac{1}{x} \right )}{\left ( x+\frac{1}{x} \right )^2}=-\frac{1}{x+\frac{1}{x}}+c=-\frac{x}{x^2+1}+c\)
Câu 8)
\(I=\int \ln \left(\frac{x+1}{x-1}\right)dx=\int \ln (x+1)dx-\int \ln (x-1)dx\)
\(\Leftrightarrow I=\int \ln (x+1)d(x+1)-\int \ln (x-1)d(x-1)\)
Xét \(\int \ln tdt\) ta có:
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t+c\)
\(\Rightarrow I=(x+1)\ln (x+1)-(x+1)-(x-1)\ln (x-1)+x-1+c\)
\(\Leftrightarrow I=(x+1)\ln(x+1)-(x-1)\ln(x-1)+c\)
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Xem thêm câu trả lời
1)\(\int_1^e\left(\frac{lnx}{x}\right)^2dx\)
2)\(\int_0^{\frac{\pi}{4}}\frac{x}{1+cos2x}dx\)
3)\(\int_0^{\frac{\pi}{4}}\frac{ln\left(cosx\right)}{cos^2x}dx\)
Câu 1)
Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=\frac{1}{x^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2\ln x}{x}\\ v=\frac{-1}{x}\end{matrix}\right.\)
\(\int \left ( \frac{\ln}{x} \right )^2dx=\frac{-\ln^2x}{x}+2\int \frac{\ln x}{x^2}dx\)
Đặt \(\left\{\begin{matrix} t=\ln x\\ dk=\frac{1}{x^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dt=\frac{1}{x}dx\\ k=-\frac{1}{x}\end{matrix}\right.\Rightarrow \int \frac{\ln x}{x^2}dx=-\frac{\ln x}{x}+\int \frac{1}{x^2}dx=\frac{-\ln x}{x}-\frac{1}{x}\)
\(\Rightarrow I=\left.\begin{matrix} e\\ 1\end{matrix}\right|\left(\frac{-\ln^2 x}{x}-\frac{2\ln x}{x}-\frac{2}{x}\right)=2-\frac{5}{e}\)
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Câu 2)
\(I=\int ^{\frac{\pi}{4}}_{0}\frac{x}{1+\cos 2x}dx=\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{x}{\cos^2x}dx\)
Đặt \(\left\{\begin{matrix} u=x\\ dv=\frac{dx}{\cos^2x}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\tan x\end{matrix}\right.\Rightarrow I=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{x\tan x}{2}-\frac{1}{2}\int^{\frac{\pi}{4}}_{0} \tan xdx\)
\(=\frac{\pi}{8}+\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{d(\cos x)}{\cos x}=\frac{\pi}{8}+\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{\ln |\cos x|}{2}=\frac{\pi}{8}+\frac{\ln\frac{\sqrt{2}}{2}}{2}\)
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Câu 3)
Đặt \(\left\{\begin{matrix} u=\ln (\cos x)\\ dv=\frac{dx}{\cos^2x}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{-\sin x}{\cos x}dx=-\tan xdx\\ v=\tan x\end{matrix}\right.\)
\(\Rightarrow I=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\tan x\ln (\cos x)+\int ^{\frac{\pi}{4}}_{0}\tan^2xdx=\ln \frac{\sqrt{2}}{2}+\int ^{\frac{\pi}{4}}_{0}(\frac{1}{\cos^2x}-1)dx\)
\(=\ln\frac{\sqrt{2}}{2}+\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|(\tan x-x)=\ln \frac{\sqrt{2}}{2}-\frac{\pi}{4}+1\)
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Giúp mình với ạ. Giải pt:
1) \(sin^2x\left(x+\frac{\pi}{4}\right)=\sqrt{2}s\text{inx}\)
2) \(3\sqrt{2}c\text{os}x-s\text{inx}=c\text{os}3x+3\sqrt{2}sinxsin2x\:\)
\(\frac{sin\left(2x+\frac{3\pi}{4}\right)}{c\text{os}\left(x+\frac{\pi}{4}\right)}+1=0\)
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\(\Leftrightarrow sin\left(2x+\frac{3\pi}{4}\right)+cos\left(x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-sin\left(2x+\frac{3\pi}{4}\right)\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=cos\left(2x+\frac{5\pi}{4}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x+\frac{5\pi}{4}=x+\frac{\pi}{4}+k2\pi\\2x+\frac{5\pi}{4}=-x-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\pi+k2\pi\\x=-\frac{\pi}{2}+\frac{k2\pi}{3}\end{matrix}\right.\)
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