Đặt \(\sqrt{5-x}=t\Rightarrow5-x=t^2\Rightarrow dx=-2t.dt\)
\(\left\{{}\begin{matrix}x=-4\Rightarrow t=3\\x=4\Rightarrow t=1\end{matrix}\right.\)
\(I=\int\limits^1_3\dfrac{-2t.dt}{1+t}=\int\limits^3_1\left(2-\dfrac{2}{t+1}\right)dt=\left(2t-2ln\left|t+1\right|\right)|^3_1=4-2ln2\)
\(\Rightarrow ab=8\)
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) \(\int\limits^4_0\dfrac{4x-1}{\sqrt{2x+1}+2}dx\)
Tính các tích phân sau bằng phương pháp đổi biến số :
a) \(\int\limits^2_1x\left(1-x\right)^5dx\) (đặt \(t=1-x\))
b) \(\int\limits^{\ln2}_0\sqrt{e^x-1}dx\) (đặt \(t=\sqrt{e^x-1}\))
c) \(\int\limits^9_1x\sqrt[3]{1-x}dx\) (đặt \(t=\sqrt[3]{1-x}\) )
d) \(\int\limits^1_{-1}\dfrac{2x+1}{\sqrt{x^2+x+1}}dx\) (đặt \(u=\sqrt{x^2+x+1}\))
e) \(\int\limits^2_1\dfrac{\sqrt{1+x^2}}{x^4}dx\) (đặt \(t=\dfrac{1}{x}\) )
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 :
a) \(\int\limits^{\dfrac{1}{2}}_{-\dfrac{1}{2}}\sqrt[3]{\left(1-x\right)^2dx}\)
b) \(\int\limits^{\dfrac{\pi}{2}}_0\sin\left(\dfrac{\pi}{4}-x\right)dx\)
c) \(\int\limits^2_{\dfrac{1}{2}}\dfrac{1}{x\left(x+1\right)}dx\)
d) \(\int\limits^2_0x\left(x+1\right)^2dx\)
e) \(\int\limits^2_{\dfrac{1}{2}}\dfrac{1-3x}{\left(x+1\right)^2}dx\)
g) \(\int\limits^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}\sin3x\cos5xdx\)
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) \(\int\limits^2_0\left(\dfrac{1}{1+x+x^2+x^3}+1\right)dx=0\)
Sử dụng phương pháp đổi biến số, hãy tính :
a) \(\int\limits^3_0\dfrac{x^2}{\left(1+x\right)^{\dfrac{3}{2}}}dx\) (đặt \(u=x+1\))
b) \(\int\limits^1_0\sqrt{1-x^2}dx\) (đặt \(x=\sin t\))
c) \(\int\limits^1_0\dfrac{e^x\left(1+x\right)}{1+xe^x}dx\) (đặt \(u=1+xe^x\))
d) \(\int\limits^{\dfrac{a}{2}}_0\dfrac{1}{\sqrt{a^2-x^2}}dx\) (\(a>0\)) (đặt \(x=a\sin t\))
chứng minh:
\(\int\limits^1_0\dfrac{ln\left(x+\sqrt{1-x^2}\right)}{x}dx=\dfrac{3}{4}\int\limits\dfrac{ln\left(1+x\right)}{x}^1_0dx\)
Áp dụng phương pháp tính tích phân, hãy tính các tích phân sau :
a) \(\int\limits^{\dfrac{\pi}{2}}_0x\cos2xdx\)
b) \(\int\limits^{\ln2}_0xe^{-2x}dx\)
c) \(\int\limits^1_0\ln\left(2x+1\right)dx\)
d) \(\int\limits^3_2\left|\ln\left(x-1\right)-\ln\left(x+1\right)\right|dx\)
e) \(\int\limits^2_{\dfrac{1}{2}}\left(1+x-\dfrac{1}{x}\right)e^{x+\dfrac{1}{x}}dx\)
g) \(\int\limits^{\dfrac{\pi}{2}}_0x\cos x\sin^2xdx\)
h) \(\int\limits^1_0\dfrac{xe^x}{\left(1+x\right)^2}dx\)
i) \(\int\limits^e_1\dfrac{1+x\ln x}{x}e^xdx\)
1) \(\int\limits^2_1\dfrac{\sqrt{x^2-1}}{x}dx\)
2) \(\int x.\sqrt{1-x^4}dx\)
3)\(\int\dfrac{1}{x^2.\sqrt{25-x^2}}dx\)
4) \(\int\dfrac{\sqrt{x^2-9}}{x^3}dx\)
1, I = \(\int\limits^1_0\dfrac{2x+1}{x^2+x+1}dx\)
2,\(\int\limits^{\dfrac{1}{2}}_0\dfrac{5xdx}{\left(1-x^2\right)^3}\)
3, \(\int\limits^1_0\dfrac{2x}{\left(x+1\right)^3}dx\)
4, \(\int\limits^1_0\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}dx\)
5, \(\int\limits^1_0\dfrac{x^2dx}{x^6-9}\)
6, \(\int\limits^2_1\dfrac{2x-1}{x^2\left(x+1\right)}dx\)
