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.\(\int_0^{\dfrac{\pi}{4}}\dfrac{\sin2x}{\sqrt{1+\cos^4x}}dx\)
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.\(\int\limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}\sin x\sqrt{3+\cos^6x}dx\)
Giải các pt
a) \(\sqrt{2}\sin\left(2x+\dfrac{\pi}{4}\right)=3\sin x+\cos x+2\)
b) \(\dfrac{\left(2-\sqrt{3}\right)\cos x-2\sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2\cos x-1}=1\)
c) \(2\sqrt{2}\cos\left(\dfrac{5\pi}{12}-x\right)\sin x=1\)
a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
c.
\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)
\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)
\(\Leftrightarrow...\)
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
2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)
\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)
\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)
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=...
Hãy chỉ ra kết quả nào dưới đây đúng :
a) \(\int\limits^{\dfrac{\pi}{2}}_0\sin xdx+\int\limits^{\dfrac{3\pi}{2}}_{\dfrac{\pi}{2}}\sin xdx+\int\limits^{2\pi}_{\dfrac{3\pi}{2}}\sin xdx=0\)
b) \(\int\limits^{\dfrac{\pi}{2}}_0\left(\sqrt[3]{\sin x}-\sqrt[3]{\cos x}\right)dx=0\)
c) \(\int\limits^{\dfrac{1}{2}}_{-\dfrac{1}{2}}\ln\dfrac{1-x}{1+x}dx=0\)
d) \(\int\limits^2_0\left(\dfrac{1}{1+x+x^2+x^3}+1\right)dx=0\)
giải phương trình
a) \(sin\left(x-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
b) \(cos\left(x+\dfrac{\pi}{4}\right)=cos\dfrac{3\pi}{4}\)
c) \(tan2x=tan\left(x+\dfrac{\pi}{3}\right)\)
d) \(cot2x=-\dfrac{\sqrt{3}}{3}\)
a: \(sin\left(x-\dfrac{\Omega}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
=>\(sin\left(x-\dfrac{\Omega}{4}\right)=sin\left(-\dfrac{\Omega}{4}\right)\)
=>\(\left[{}\begin{matrix}x-\dfrac{\Omega}{4}=-\dfrac{\Omega}{4}+k2\Omega\\x-\dfrac{\Omega}{4}=\Omega+\dfrac{\Omega}{4}+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Omega\\x=\dfrac{3}{2}\Omega+k2\Omega\end{matrix}\right.\)
b: \(cos\left(x+\dfrac{\Omega}{4}\right)=cos\left(\dfrac{3}{4}\Omega\right)\)
=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{4}=\dfrac{3}{4}\Omega+k2\Omega\\x+\dfrac{\Omega}{4}=-\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\Omega+k2\Omega\\x=-\Omega+k2\Omega\end{matrix}\right.\)
c: ĐKXĐ: \(\left\{{}\begin{matrix}2x< >\dfrac{\Omega}{2}+k\Omega\\x+\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\\x< >\dfrac{1}{6}\Omega+k\Omega\end{matrix}\right.\)
\(tan2x=tan\left(x+\dfrac{\Omega}{3}\right)\)
=>\(2x=x+\dfrac{\Omega}{3}+k\Omega\)
=>\(x=\dfrac{\Omega}{3}+k\Omega\)
d: ĐKXĐ: \(2x< >k\Omega\)
=>\(x< >\dfrac{k\Omega}{2}\)
\(cot2x=-\dfrac{\sqrt{3}}{3}\)
=>\(cot2x=cot\left(-\dfrac{\Omega}{3}\right)\)
=>\(2x=-\dfrac{\Omega}{3}+k\Omega\)
=>\(x=-\dfrac{\Omega}{6}+\dfrac{k\Omega}{2}\)
Tìm đạo hàm các hàm số:
1, \(y=\tan(3x-\dfrac{\pi}{4})+\cot(2x-\dfrac{\pi}{3})+\cos(x+\dfrac{\pi}{6})\)
2, \(y=\dfrac{\sqrt{\sin x+2}}{2x+1}\)
3, \(y=\cos(3x+\dfrac{\pi}{3})-\sin(2x+\dfrac{\pi}{6})+\cot(x+\dfrac{\pi}{4})\)
a.
\(y'=\dfrac{3}{cos^2\left(3x-\dfrac{\pi}{4}\right)}-\dfrac{2}{sin^2\left(2x-\dfrac{\pi}{3}\right)}-sin\left(x+\dfrac{\pi}{6}\right)\)
b.
\(y'=\dfrac{\dfrac{\left(2x+1\right)cosx}{2\sqrt{sinx+2}}-2\sqrt{sinx+2}}{\left(2x+1\right)^2}=\dfrac{\left(2x+1\right)cosx-4\left(sinx+2\right)}{\left(2x+1\right)^2}\)
c.
\(y'=-3sin\left(3x+\dfrac{\pi}{3}\right)-2cos\left(2x+\dfrac{\pi}{6}\right)-\dfrac{1}{sin^2\left(x+\dfrac{\pi}{4}\right)}\)
1) sin\(\sin\left[\pi sin2x\right]\)=1
2) cos\(\left[\dfrac{\pi}{2}.cos\left(x-\dfrac{\pi}{4}\right)\right]\)=\(\dfrac{\sqrt{2}}{2}\)
3) sin(x+24*) + sin(x+144*) = cos20*
1.
Chắc đề là \(sin\left[\pi sin2x\right]=1?\)
\(\Leftrightarrow\pi.sin2x=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow sin2x=\dfrac{1}{2}+2k\) (1)
Do \(-1\le sin2x\le1\Rightarrow-1\le\dfrac{1}{2}+2k\le1\)
\(\Rightarrow-\dfrac{3}{4}\le k\le\dfrac{1}{4}\Rightarrow k=0\)
Thế vào (1)
\(\Rightarrow sin2x=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{6}+n2\pi\\2x=\dfrac{5\pi}{6}+m2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+n\pi\\x=\dfrac{5\pi}{12}+m\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+k2\pi\\\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{\pi}{4}+k_12\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}+4k\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}+4k_1\end{matrix}\right.\) (2)
Do \(-1\le cos\left(x-\dfrac{\pi}{4}\right)\le1\Rightarrow\left\{{}\begin{matrix}-1\le\dfrac{1}{2}+4k\le1\\-1\le-\dfrac{1}{2}+4k_1\le1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k=0\\k_1=0\end{matrix}\right.\)
Thế vào (2):
\(\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\) chắc bạn tự giải tiếp được
3.
\(\Leftrightarrow2sin\left(x+84^0\right).cos\left(60^0\right)=cos20^0\)
\(\Leftrightarrow sin\left(x+84^0\right)=sin70^0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+84^0=70^0+k360^0\\x+84^0=110^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-14^0+k360^0\\x=26^0+k360^0\end{matrix}\right.\)
Giải PT
a1) \(3.\cos4x-2^{ }\cos^23x=1\)
a2) \(2\cos2x-8\cos x+7=\dfrac{1}{\cos x}\)
a3) \(\dfrac{\left(1+\sin x+\cos2x\right)\sin\left(x+\dfrac{\pi}{4}\right)}{1+\tan x}=\dfrac{1}{\sqrt{2}}\cos x\)
a4) \(9\sin x+6\cos x-3\sin2x+\cos2x=8\)
a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)
\(\Leftrightarrow3cos4x-cos6x-2=0\)
Đặt \(t=2x\)
Pttt:\(3cos2t-cos3t-2=0\)
\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)
\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))
\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)
\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)
\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))
Vậy...
a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)
Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)
\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)
\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))
\(\Rightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)
Vậy...
1) sin2x + 2cosx = 0
2) sin(2x -10*) = \(\dfrac{1}{2}\) (-120* <x< 90*)
3) cos(2x+10*)= \(\dfrac{\sqrt{2}}{2}\)(-180*<x<180*)
4) \(\sin^2\left(5x+\dfrac{2\pi}{5}\right)-\cos^2\)(\(\dfrac{x}{4}-\pi\)) =0
1.
\(\Leftrightarrow2sinx.cosx+2cosx=0\)
\(\Leftrightarrow2cosx\left(sinx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow cosx=0\) (do \(cosx=0\Leftrightarrow sinx=\pm1\) bao hàm luôn cả pt \(sinx=-1\))
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
2.
\(\Leftrightarrow\left[{}\begin{matrix}2x-10^0=60^0+k360^0\\2x-10^0=120^0+n360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=35^0+k180^0\\x=65^0+n180^0\end{matrix}\right.\)
Do \(-120^0< x< 90^0\Rightarrow\left\{{}\begin{matrix}-120^0< 35^0+k180^0< 90^0\\-120^0< 65^0+n180^0< 90^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k=0\\n=\left\{-1;0\right\}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=35^0\\x=-115^0\\x=65^0\end{matrix}\right.\)
3. Làm tương tự câu 2
4.
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos\left(10x+\dfrac{4\pi}{5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{x}{2}-2\pi\right)\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}-2\pi\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)=-cos\left(\dfrac{x}{2}\right)=cos\left(\pi-\dfrac{x}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}10x+\dfrac{4\pi}{5}=\pi-\dfrac{x}{2}+k2\pi\\10x+\dfrac{4\pi}{5}=\dfrac{x}{2}-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
Tìm m để hàm số \(y=\sqrt{\dfrac{m-\sin x-\cos x-2\sin x\cos x}{\sin^{2017}x-\cos^{2019}x+\sqrt{2}}}\) xác định với mọi \(x\in[-\dfrac{\pi}{2};\dfrac{\pi}{2}]\)