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 các nguyên hàm sau :
a) \(\int x\left(3-x\right)^5dx\)
b) \(\int\left(2^x-3^x\right)^2dx\)
c) \(\int x\sqrt{2-5x}dx\)
d) \(\int\dfrac{\ln\left(\cos x\right)}{\cos^2x}dx\)
e) \(\int\dfrac{x}{\sin^2x}dx\)
\(\int\dfrac{x+1}{\left(x-2\right)\left(x+3\right)}dx\)
h) \(\int\dfrac{1}{1-\sqrt{x}}dx\)
i) \(\int\sin3x\cos2xdx\)
k) \(\int\dfrac{\sin^3x}{\cos^2x}dx\)
l) \(\int\dfrac{\sin x\cos x}{\sqrt{a^2\sin^2x+b^2\cos^2x}}dx\) (\(a^2\ne b^2\))
1, \(\int\dfrac{x^3dx}{\left(x^8-4\right)^2}\)
2, \(\int\dfrac{2x+1}{x^4+2x^2+3x^2+2x-3}dx\)
3, \(\int\dfrac{sin\sqrt[3]{x}}{\sqrt[3]{x^2}}dx\)
4, \(\int\dfrac{dx}{sin^2x+2cos^2x}\)
5, \(\int\dfrac{sinx+cosx}{3+sin2x}dx\)
\(I=\int\dfrac{x^3dx}{\left(x^8-4\right)^2}\)
Đặt \(x^4=t\Rightarrow x^3dx=\dfrac{1}{4}dt\Rightarrow I=\dfrac{1}{4}\int\dfrac{dt}{\left(t^2-2\right)^2}=\dfrac{1}{4}\int\dfrac{dt}{\left(t-\sqrt{2}\right)^2\left(t+\sqrt{2}\right)^2}\)
\(=\dfrac{1}{32}\int\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)^2dt=\dfrac{1}{32}\int\left(\dfrac{1}{\left(t-\sqrt{2}\right)^2}+\dfrac{1}{\left(t+\sqrt{2}\right)^2}-\dfrac{2}{\left(t+\sqrt{2}\right)\left(t-\sqrt{2}\right)}\right)dt\)
\(=\dfrac{1}{32}\int\left(\dfrac{1}{\left(t-\sqrt{2}\right)^2}+\dfrac{1}{\left(t+\sqrt{2}\right)^2}-\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)\right)dt\)
\(=\dfrac{1}{32}\left(\dfrac{-1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}-\dfrac{1}{\sqrt{2}}ln\left|\dfrac{t-\sqrt{2}}{t+\sqrt{2}}\right|\right)+C\)
\(=\dfrac{1}{32}\left(\dfrac{-1}{x^4-\sqrt{2}}-\dfrac{1}{x^4+\sqrt{2}}-\dfrac{1}{\sqrt{2}}ln\left|\dfrac{x^4-\sqrt{2}}{x^4+\sqrt{2}}\right|\right)+C\)
2/ \(I=\int\dfrac{\left(2x+1\right)dx}{\left(x^2+x-1\right)\left(x^2+x+3\right)}=\dfrac{1}{4}\int\left(\dfrac{1}{x^2+x-1}-\dfrac{1}{x^2+x+3}\right)\left(2x+1\right)dx\)
\(=\dfrac{1}{4}\int\left(\dfrac{2x+1}{x^2+x-1}-\dfrac{2x+1}{x^2+x+3}\right)dx\)
\(=\dfrac{1}{4}\left(\int\dfrac{d\left(x^2+x-1\right)}{x^2+x-1}-\int\dfrac{d\left(x^2+x+3\right)}{x^2+x+3}\right)\)
\(=\dfrac{1}{4}ln\left|\dfrac{x^2+x-1}{x^2+x+3}\right|+C\)
3/ Đặt \(\sqrt[3]{x}=t\Rightarrow x=t^3\Rightarrow dx=3t^2dt\)
\(\Rightarrow I=\int\dfrac{3t^2.sint.dt}{t^2}=3\int sint.dt=-3cost+C=-3cos\left(\sqrt[3]{x}\right)+C\)
4/ \(I=\int\dfrac{dx}{1+cos^2x}=\int\dfrac{\dfrac{1}{cos^2x}dx}{\dfrac{1}{cos^2x}+1}\)
Đặt \(t=tanx\Rightarrow\left\{{}\begin{matrix}dt=\dfrac{1}{cos^2x}dx\\\dfrac{1}{cos^2x}=1+tan^2x=1+t^2\end{matrix}\right.\)
\(\Rightarrow I=\int\dfrac{dt}{1+t^2+1}=\int\dfrac{dt}{t^2+2}=\dfrac{1}{2}\int\dfrac{dt}{\left(\dfrac{t}{\sqrt{2}}\right)^2+1}\)
\(=\dfrac{1}{2}.\sqrt{2}.arctan\left(\dfrac{t}{\sqrt{2}}\right)+C=\dfrac{1}{\sqrt{2}}arctan\left(\dfrac{tanx}{\sqrt{2}}\right)+C\)
5/ \(I=\int\dfrac{sinx+cosx}{4+2sinx.cosx-sin^2x-cos^2x}dx=\int\dfrac{sinx+cosx}{4-\left(sinx-cosx\right)^2}dx\)
Đặt \(sinx-cosx=t\Rightarrow\left(cosx+sinx\right)dx=dt\)
\(\Rightarrow I=\int\dfrac{dt}{4-t^2}=-\int\dfrac{dt}{\left(t-2\right)\left(t+2\right)}=\dfrac{1}{4}\int\left(\dfrac{1}{t+2}-\dfrac{1}{t-2}\right)dt\)
\(=\dfrac{1}{4}ln\left|\dfrac{t+2}{t-2}\right|+C=\dfrac{1}{4}ln\left|\dfrac{sinx-cosx+2}{sinx-cosx-2}\right|+C\)
Ơ bài 1 nhầm số 4 thành số 2 rồi, bạn sửa lại 1 chút nhé :D
Còn 1 cách làm khác nữa là lượng giác hóa
Đặt \(x^4=2sint\Rightarrow x^3dx=\dfrac{1}{2}cost.dt\)
\(\Rightarrow I=\dfrac{1}{2}\int\dfrac{cost.dt}{\left(4sin^2t-4\right)^2}=\dfrac{1}{32}\int\dfrac{cost.dt}{cos^4t}=\dfrac{1}{32}\int\dfrac{dt}{cos^3t}\)
Đặt \(\left\{{}\begin{matrix}u=\dfrac{1}{cost}\\dv=\dfrac{dt}{cos^2t}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{sint.dt}{cos^2t}\\v=tant\end{matrix}\right.\)
\(\Rightarrow32I=\dfrac{tant}{cost}-\int\dfrac{tant.sint.dt}{cos^2t}=\dfrac{sint}{cos^2t}-\int\dfrac{sin^2t.dt}{cos^3t}\)
\(=\dfrac{sint}{1-sin^2t}-\int\dfrac{1-cos^2t}{cos^3t}dt=\dfrac{sint}{1-sin^2t}-\int\dfrac{dt}{cos^3t}+\int\dfrac{1}{cosx}dx\)
Chú ý rằng \(\int\dfrac{dt}{cos^3t}=32I\)
\(\Rightarrow32I=\dfrac{sint}{1-sin^2t}-32I+\int\dfrac{cost.dt}{cos^2t}\)
\(\Rightarrow64I=\dfrac{sint}{1-sin^2t}-\int\dfrac{d\left(sint\right)}{sin^2t-1}=\dfrac{sint}{1-sin^2t}-\dfrac{1}{2}ln\left|\dfrac{sint-1}{sint+1}\right|+C\)
\(\Rightarrow I=\dfrac{1}{64}\left(\dfrac{2x^4}{4-x^8}-\dfrac{1}{2}ln\left|\dfrac{x^4-2}{x^4+2}\right|\right)+C\)
Tính các tích phân sau :
a) \(\int\limits^1_0\left(y^3+3y^2-2\right)dy\)
b) \(\int\limits^4_1\left(t+\dfrac{1}{\sqrt{t}}-\dfrac{1}{t^2}\right)dt\)
c) \(\int\limits^{\dfrac{\pi}{2}}_0\left(2\cos x-\sin2x\right)dx\)
d) \(\int\limits^1_0\left(3^s-2^s\right)^2ds\)
e) \(\int\limits^{\dfrac{\pi}{3}}_0\cos3xdx+\int\limits^{\dfrac{3\pi}{2}}_0\cos3xdx+\int\limits^{\dfrac{5\pi}{2}}_{\dfrac{3\pi}{2}}\cos3xdx\)
g) \(\int\limits^3_0\left|x^2-x-2\right|dx\)
h) \(\int\limits^{\dfrac{5\pi}{4}}_{\pi}\dfrac{\sin x-\cos x}{\sqrt{1+\sin2x}}dx\)
i) \(\int\limits^4_0\dfrac{4x-1}{\sqrt{2x+1}+2}dx\)
Câu nào mình biết thì mình làm nha.
1) Đổi thành \(\dfrac{y^4}{4}+y^3-2y\) rồi thế số.KQ là \(\dfrac{-3}{4}\)
2) Biến đổi thành \(\dfrac{t^2}{2}+2\sqrt{t}+\dfrac{1}{t}\) và thế số.KQ là \(\dfrac{35}{4}\)
3) Biến đổi thành 2sinx + cos(2x)/2 và thế số.KQ là 1
Tính các tích phân sau :
a) \(\int\limits^2_0\left|1-x\right|dx\)
b) \(\int\limits^{\dfrac{\pi}{2}}_0\sin^2xdx\)
c) \(\int\limits^{ln2}_0\dfrac{e^{2x+1}+1}{e^x}dx\)
d) \(\int\limits^{\pi}_0\sin2x\cos^2xdx\)
Tính các nguyên hàm sau đây :
a) \(\int\left(x+\ln x\right)x^2dx\)
b) \(\int\left(x+\sin^2x\right)\sin xdx\)
c) \(\int\left(x+e^x\right)e^{2x}dx\)
d) \(\int\left(x+\sin x\right)\dfrac{dx}{\cos^2x}\)
e) \(\int\dfrac{e^x\cos x+\left(e^x+1\right)\sin x}{e^x\sin x}dx\)
a) \(\int\left(x+\ln x\right)x^2\text{d}x=\int x^3\text{d}x+\int x^2\ln x\text{dx}\)
\(=\dfrac{x^4}{4}+\int x^2\ln x\text{dx}+C\) (*)
Để tính: \(\int x^2\ln x\text{dx}\) ta sử dụng công thức tính tích phân từng phần như sau:
Đặt \(\left\{{}\begin{matrix}u=\ln x\\v'=x^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=\dfrac{1}{x}\\v=\dfrac{1}{3}x^3\end{matrix}\right.\)
Suy ra:
\(\int x^2\ln x\text{dx}=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}\int x^2\text{dx}\)
\(=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}.\dfrac{1}{3}x^3\)
Thay vào (*) ta tính được nguyên hàm của hàm số đã cho bằng:
(*) \(=\dfrac{1}{3}x^3-\dfrac{1}{3}x^3\ln x+\dfrac{1}{9}x^3+C\)
\(=\dfrac{4}{9}x^3-\dfrac{1}{3}x^3\ln x+C\)
b) Đặt \(\left\{{}\begin{matrix}u=x+\sin^2x\\v'=\sin x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u'=1+2\sin x.\cos x\\v=-\cos x\end{matrix}\right.\)
Ta có:
\(\int\left(x+\sin^2x\right)\sin x\text{dx}=-\left(x+\sin^2x\right)\cos x+\int\left(1+2\sin x\cos^2x\right)\text{dx}\)
\(=-\left(x+\sin^2x\right)\cos x+\int\cos x\text{dx}+2\int\sin x.\cos^2x\text{dx}\)
\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\int\cos^2x.d\left(\cos x\right)\)
\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\dfrac{\cos^3x}{3}+C\)
c) Đặt \(\left\{{}\begin{matrix}u=x+e^x\\v'=e^{2x}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=1+e^x\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)
Ta có:
\(\int\left(x+e^x\right)e^{2x}\text{dx}=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}\int\left(1+e^x\right)e^{2x}\text{dx}\)
\(=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}\int e^{2x}\text{dx}-\dfrac{1}{2}\int e^{3x}\text{dx}\)
\(=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}.\dfrac{1}{2}e^{2x}-\dfrac{1}{2}.\dfrac{1}{3}e^{3x}\)
\(=\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}+\dfrac{1}{3}e^{3x}\)
Bằng cách biến đổi các hàm số lượng giác, hãy tính :
a) \(\int\sin^4xdx\)
b) \(\int\dfrac{1}{\sin^3x}dx\)
c) \(\int\sin^3x\cos^4xdx\)
d) \(\int\sin^4x\cos^4xdx\)
e) \(\int\dfrac{1}{\cos x\sin^2x}dx\)
g) \(\int\dfrac{1+\sin x}{1+\cos x}dx\)
a) \(\sin^4x=\left(\sin^2x\right)^2=\left(\dfrac{1-\cos2x}{2}\right)^2\)
\(=\dfrac{1}{4}\left(1-2\cos2x+\cos^22x\right)\)
\(=\dfrac{1}{4}\left(1-2.\cos2x+\dfrac{1+\cos4x}{2}\right)\)
\(=\dfrac{3}{8}-\dfrac{1}{2}\cos2x+\dfrac{1}{8}\cos4x\)
Vậy:
\(\int\sin^4x\text{dx}=\int\left(\dfrac{3}{8}-\dfrac{1}{2}\cos2x+\dfrac{1}{8}\cos4x\right)\text{dx}\)
\(=\dfrac{3}{8}x-\dfrac{1}{4}\sin2x+\dfrac{1}{32}\sin4x+C\)
Tính :
a) \(\int\limits^2_{-1}\left(5x^2-x+e^{0,5x}\right)dx\)
b) \(\int\limits^2_{0,5}\left(2\sqrt{x}+\dfrac{3}{x^2}+\cos x\right)dx\)
c) \(\int\limits^2_1\dfrac{dx}{\sqrt{2x+3}}\) (đặt \(t=\sqrt{2x+3}\) )
d) \(\int\limits^2_1\sqrt[3]{3x^3+4}x^2dx\) (đặt \(t=\sqrt[3]{3x^3+4}\) )
e) \(\int\limits^2_{-2}\left(x-2\right)\left|x\right|dx\)
g) \(\int\limits^0_1x\cos xdx\)
h) \(\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{1+\sin2x+\cos2x}{\sin x+\cos x}dx\)
i) \(\int\limits^{\dfrac{\pi}{2}}_0e^x\sin xdx\)
k) \(\int\limits^e_1x^2\ln^2xdx\)
a) \(\int sin2x.cosxdx\)
b) \(\int tanxdx\)
c) \(\int\dfrac{sinx}{1+3cosx}dx\)
d) \(\int sin^3xdx\)
e) \(\int sin^2xdx\)
f) \(\int cos^23x\)
g) \(f\left(x\right)=\dfrac{1}{sin^2x.cos^2x}\)
h) \(f\left(x\right)=\dfrac{cos2x}{sin^2x.cos^2x}\)
i) \(\int2sin3x.cos2xdx\)
j) \(\int e^x\left(2+\dfrac{e^{-x}}{cos^2x}\right)dx\)
\(a,\int sin2x.cosxdx=\int\dfrac{1}{2}\left[sin3x+sinx\right]dx=\dfrac{1}{2}\int sin3xdx+\dfrac{1}{2}\int sinxdx=\dfrac{-1}{6}cos3x-\dfrac{1}{2}cosx\)
phần a bạn thêm +C vào đáp án nhé
\(i,\int2sinx3x.cos2xdx=2\int\dfrac{1}{2}\left(sin5x+sinx\right)dx=\int sin5xdx+\int sinxdx=-\dfrac{1}{5}cos5x-cosx+C\)
\(g,\int\dfrac{1}{sin^2x.cos^2x}=\int\dfrac{sin^2x+cos^2x}{sin^2x.cos^2x}=\int\dfrac{1}{cos^2x}dx+\int\dfrac{1}{sin^2x}dx=tanx-cotx+C\)
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.\(\int\limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}\sin x\sqrt{3+\cos^6x}dx\)
