tính các tích phân
1.\(\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}e^{\sin x}\cos xdx\)
2.\(\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}e^{2\cos x+1}\sin xdx\)
3,\(\int_1^e\dfrac{e^{2lnx+1}}{x}dx\)
4.\(\int_0^1xe^{x^2+2}dx\)
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=...
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.\(\int\limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}\sin x\sqrt{3+\cos^6x}dx\)
Tính các tích phân sau bằng phương pháp tính tích phân từng phần :
a) \(\int\limits^{e^4}_1\sqrt{x}\ln xdx\)
b) \(\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{xdx}{\sin^2x}\)
c) \(\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx\)
d) \(\int\limits^0_{-1}\left(2x+3\right)e^{-x}dx\)
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) \(\int\limits^2_0\left(\dfrac{1}{1+x+x^2+x^3}+1\right)dx=0\)
Tính các tích phân sau :
a) \(\int\limits^{\dfrac{\pi}{4}}_0\cos2x.\cos^2xdx\)
b) \(\int\limits^1_{\dfrac{1}{2}}\dfrac{e^x}{e^{2x}-1}dx\)
c) \(\int\limits^1_0\dfrac{x+2}{x^2+2x+1}\ln\left(x+1\right)dx\)
d) \(\int\limits^{\dfrac{\pi}{4}}_0\dfrac{x\sin x+\left(x+1\right)\cos x}{x\sin x+\cos x}dx\)
a)
Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)
\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)
\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)
b)
\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)
\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)
c)
Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).
Đặt \(x+1=t\)
\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
d)
\(D=\int ^{\frac{\pi}{4}}_{0}\frac{x\sin x+(x+1)\cos x}{x\sin x+\cos x}dx=\int ^{\frac{\pi}{4}}_{0}dx+\int ^{\frac{\pi}{4}}_{0}\frac{x\cos x}{x\sin x+\cos x}dx\)
Ta có:
\(\int ^{\frac{\pi}{4}}_{0}dx=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|x=\frac{\pi}{4}\)
\(\int ^{\frac{\pi}{4}}_{0}\frac{x\cos xdx}{x\sin x+\cos x}=\int ^{\frac{\pi}{4}}_{0}\frac{d(x\sin x+\cos x)}{x\sin x+\cos x}=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\ln |x\sin x+\cos x|\)
\(=\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)
Suy ra \(D=\frac{\pi}{4}+\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)
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:
\(\int ^{1}_{0}\frac{(x+1)^2}{x^2+1}dx=\left.\begin{matrix} 1\\ 0\end{matrix}\right|(x+\ln (x^2+1)+c)=\ln 2+1\)
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: \(\int 3x(x+\sqrt{x^2+16})dx=x^3+\sqrt{(x^2+16)^3}+c\)
\(\Rightarrow \int ^{3}_{0}3x(x+\sqrt{x^2+16})dx=\left.\begin{matrix} 3\\ 0\end{matrix}\right|(x^3+\sqrt{(x^2+16)^3}+c)=88\)
a) \(\int_{\dfrac{\pi}{8}}^{\dfrac{2\pi}{8}}\)\(\dfrac{dx}{sin^2xcos^2x}\)
b) \(\int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}}\)\(\dfrac{cos2xdx}{sin^2xcos^2x}\)
c) \(\int_0^{\dfrac{\pi}{3}}\)\(\dfrac{cos3x}{cosx}\)dx
\(\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{dx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{2d\left(2x\right)}{sin^22x}=-2cot2x|^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}=...\)
\(\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos2xdx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos^2x-sin^2x}{sin^2x.cos^2x}dx=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\left(\dfrac{1}{sin^2x}-\dfrac{1}{cos^2x}\right)dx=\left(-cotx-tanx\right)|^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\)
\(\int\limits^{\dfrac{\pi}{3}}_0\dfrac{cos3x}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\dfrac{4cos^3x-3cosx}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\left(4cos^2x-3\right)dx\)
\(=\int\limits^{\dfrac{\pi}{3}}_0\left(2cos2x-1\right)dx=\left(sin2x-x\right)|^{\dfrac{\pi}{3}}_0=...\)
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.
\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)
\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)
\(\Leftrightarrow...\)
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.
\(y'=-3sin\left(3x+\dfrac{\pi}{3}\right)-2cos\left(2x+\dfrac{\pi}{6}\right)-\dfrac{1}{sin^2\left(x+\dfrac{\pi}{4}\right)}\)