1)\(\int_1^e\left(\frac{lnx}{x}\right)^2dx\)
2)\(\int_0^{\frac{\pi}{4}}\frac{x}{1+cos2x}dx\)
3)\(\int_0^{\frac{\pi}{4}}\frac{ln\left(cosx\right)}{cos^2x}dx\)
Những câu hỏi liên quan
1/ Iint_{-2}^2left|x^2-1right|dx2/ I int_1^esqrt{x}.lnxdx3/ I int_0^{dfrac{pi}{2}}left(e^{sinx}+cosxright)cosxdx4/ I int_0^{dfrac{pi}{2}}dfrac{sin2x}{sqrt{cos^2x+4sin^2x}}dx5/ I int_0^{dfrac{pi}{4}}sqrt{2}cossqrt{x}dx6/ I int_1^{sqrt{e}}dfrac{1}{xsqrt{1-ln^2x}}dx7/ I int_{-dfrac{pi}{4}}^{dfrac{pi}{4}}dfrac{sin^6x+cos^6x}{6^x+1}dx
Đọc tiếp
1/ I=\(\int_{-2}^2\left|x^2-1\right|dx\)
2/ I= \(\int_1^e\sqrt{x}.lnxdx\)
3/ I= \(\int_0^{\dfrac{\pi}{2}}\left(e^{sinx}+cosx\right)cosxdx\)
4/ I= \(\int_0^{\dfrac{pi}{2}}\dfrac{sin2x}{\sqrt{cos^2x+4sin^2x}}dx\)
5/ I= \(\int_0^{\dfrac{\pi}{4}}\sqrt{2}cos\sqrt{x}dx\)
6/ I= \(\int_1^{\sqrt{e}}\dfrac{1}{x\sqrt{1-ln^2x}}dx\)
7/ I= \(\int_{-\dfrac{\pi}{4}}^{\dfrac{\pi}{4}}\dfrac{sin^6x+cos^6x}{6^x+1}dx\)
Nhìn đề dữ dội y hệt cr của tui z :( Để làm từ từ
Lập bảng xét dấu cho \(\left|x^2-1\right|\) trên đoạn \(\left[-2;2\right]\)
| x | -2 | -1 | 1 | 2 |
| \(x^2-1\) | 0 | 0 |
\(\left(-2;-1\right):+\)
\(\left(-1;1\right):-\)
\(\left(1;2\right):+\)
\(\Rightarrow I=\int\limits^{-1}_{-2}\left|x^2-1\right|dx+\int\limits^1_{-1}\left|x^2-1\right|dx+\int\limits^2_1\left|x^2-1\right|dx\)
\(=\int\limits^{-1}_{-2}\left(x^2-1\right)dx-\int\limits^1_{-1}\left(x^2-1\right)dx+\int\limits^2_1\left(x^2-1\right)dx\)
\(=\left(\dfrac{x^3}{3}-x\right)|^{-1}_{-2}-\left(\dfrac{x^3}{3}-x\right)|^1_{-1}+\left(\dfrac{x^3}{3}-x\right)|^2_1\)
Bạn tự thay cận vô tính nhé :), hiện mình ko cầm theo máy tính
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2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)
\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)
\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)
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3/ \(I=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx+\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)
Xét \(A=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx\)
\(t=\sin x\Rightarrow dt=\cos x.dx\Rightarrow A=\int\limits^{\dfrac{\pi}{2}}_0e^t.dt=e^{\sin x}|^{\dfrac{\pi}{2}}_0\)
Xét \(B=\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)
\(=\int\limits^{\dfrac{\pi}{2}}_0\dfrac{1+\cos2x}{2}.dx=\dfrac{1}{2}.\int\limits^{\dfrac{\pi}{2}}_0dx+\dfrac{1}{2}\int\limits^{\dfrac{\pi}{2}}_0\cos2x.dx\)
\(=\dfrac{1}{2}x|^{\dfrac{\pi}{2}}_0+\dfrac{1}{2}.\dfrac{1}{2}\sin2x|^{\dfrac{\pi}{2}}_0\)
I=A+B=...
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Xem thêm câu trả lời
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
Đọc tiếp
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
\(\int_0^{\frac{\Pi}{2}}c\text{os}^2x\left(1-sin^3x\right)dx\)
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
1)
\(I=\int\left(cos^2x-cos^2x\cdot sin^3x\right)dx\\ =\int cos^2x\cdot dx-\int cos^2x\cdot sin^3x\cdot dx\\ =\frac{1}{2}\int\left(cos2x+1\right)dx+\int cos^2x\left(1-cos^2x\right)d\left(cosx\right)\\ =\frac{1}{4}sin2x+\frac{1}{2}+\frac{cos^3x}{3}-\frac{cos^5x}{5}+C\)
....
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2) Xét riêng mẫu số:
\(sin2x+2\left(1+sinx+cosx\right)\\ =\left(sin2x+1\right)+2\left(sinx+cosx\right)+1\\ =\left(sinx+cosx\right)^2+2\left(sinx+cosx\right)+1\\ =\left(sinx+cosx+1\right)^2\\ =\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2\)
Khi đó:
\(I_2=\int\frac{sin\left(x-\frac{\pi}{4}\right)}{\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2}dx\\ =-\frac{1}{\sqrt{2}}\int\frac{d\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]}{\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2}\\ =\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1}+C=\frac{1}{2cos\left(x-\frac{\pi}{4}\right)+1}\)
...
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1.\(\int_0^1x\left(e^{2x^{ }}-\frac{x^2}{\sqrt{4-x^2}}\right)dx\)
2.Giải pt \(\sin\left(2x+\frac{\pi}{2}\right)+\cos x-\sin x=0\)
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.
1.\(\int_0^{\dfrac{\pi}{4}}\dfrac{\sin2x}{\sqrt{1+\cos^4x}}dx\)
2.\(\int_0^{ln3}\dfrac{e^x}{\sqrt{e^x+1}+1}dx\)
3.\(\int_1^2\dfrac{3x+1}{\sqrt{x^2+3x+9}}dx\)
4.\(\int\limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}\sin x\sqrt{3+\cos^6x}dx\)
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)...
Đọc tiếp
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)\(\int\frac{c\text{os}x+s\text{inx}}{\left(e^xs\text{inx}+1\right)s\text{inx}}dx\)
Tính (trình bày cách giải ln nka):
a) \(\int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}}\dfrac{1}{cos^4x}dx\)
b) \(\int_0^1\dfrac{\left(x+1\right)^2}{x^2+1}dx\)
c)\(\int_1^2\dfrac{x^2+2lnx}{x}dx\)
d) \(\int_1^2\dfrac{x^2+3x+1}{x^2+x}dx\)
e) \(\int_0^33x\left(x+\sqrt{x^2+16}\right)dx\)
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:
\(\int ^{1}_{0}\frac{(x+1)^2}{x^2+1}dx=\left.\begin{matrix} 1\\ 0\end{matrix}\right|(x+\ln (x^2+1)+c)=\ln 2+1\)
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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)
\(\int^{2}_{1} \frac{x^2+3x+1}{x^2+x}dx=\int ^{2}_{1}dx+\int ^{2}_{1}\frac{2x+1}{x^2+x}dx\)
\(=\left.\begin{matrix} 2\\ 1\end{matrix}\right|x+\int ^{2}_{1}\frac{d(x^2+x)}{x^2+x}=1+\left.\begin{matrix} 2\\ 1\end{matrix}\right|\ln |x^2+x|=1+\ln 6-\ln 2\)
\(=1+\ln 3\)
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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: \(\int 3x(x+\sqrt{x^2+16})dx=x^3+\sqrt{(x^2+16)^3}+c\)
\(\Rightarrow \int ^{3}_{0}3x(x+\sqrt{x^2+16})dx=\left.\begin{matrix} 3\\ 0\end{matrix}\right|(x^3+\sqrt{(x^2+16)^3}+c)=88\)
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Cho f(x) liên tục trên R thỏa mãn int_0^{frac{1}{2}}fleft(sqrt{1-2x^2}right)dx frac{7}{6} và f (frac{1}{sqrt{2}}) 1. Tính I int_0^{frac{Pi}{4}}fleft(cosxright)sin^2xdx
A. frac{1}{2} B.frac{sqrt{2}}{3} C. frac{2sqrt{2}}{3} D. 1
Đọc tiếp
Cho f(x) liên tục trên R thỏa mãn \(\int_0^{\frac{1}{2}}f\left(\sqrt{1-2x^2}\right)dx\) = \(\frac{7}{6}\) và f (\(\frac{1}{\sqrt{2}}\)) =1. Tính I = \(\int_0^{\frac{\Pi}{4}}f'\left(cosx\right)sin^2xdx\)
A. \(\frac{1}{2}\) B.\(\frac{\sqrt{2}}{3}\) C. \(\frac{2\sqrt{2}}{3}\) D. 1
Đặt \(x=\frac{\sqrt{2}}{2}sint\Rightarrow dx=\frac{\sqrt{2}}{2}cost.dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=\frac{1}{2}\Rightarrow t=\frac{\pi}{4}\end{matrix}\right.\)
\(\int\limits^{\frac{1}{2}}_0f\left(\sqrt{1-2x^2}\right)dx=\frac{\sqrt{2}}{2}\int\limits^{\frac{\pi}{4}}_0f\left(cost\right).costdt=\frac{\sqrt{2}}{2}\int\limits^{\frac{\pi}{4}}_0f\left(cosx\right)cosxdx=\frac{7}{6}\)
\(\Rightarrow J=\int\limits^{\frac{\pi}{4}}_0f\left(cosx\right).cosx.dx=\frac{7\sqrt{2}}{6}\)
Đặt \(\left\{{}\begin{matrix}u=f\left(cosx\right)\\dv=cosx.dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-sinx.f'\left(cosx\right)dx\\v=sinx\end{matrix}\right.\)
\(\Rightarrow J=sinx.f\left(cosx\right)|^{\frac{\pi}{4}}_0+\int\limits^{\frac{\pi}{4}}_0f'\left(cosx\right)sin^2x.dx=\frac{\sqrt{2}}{2}+I\)
\(\Rightarrow I=\frac{7\sqrt{2}}{6}-\frac{\sqrt{2}}{2}=\frac{2\sqrt{2}}{3}\)
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