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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
a.
Đặt \(\left\{{}\begin{matrix}u=2-x\\dv=sinxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-dx\\v=-cosx\end{matrix}\right.\)
\(\Rightarrow I=\left(x-2\right).cosx|^{\dfrac{\pi}{2}}_0-\int\limits^{\dfrac{\pi}{2}}_0cosx.dx=2-1=1\)
b. Đặt \(cos2x=t\Rightarrow-2sin2x.dx=dt\Rightarrow sin2xdx=-\dfrac{1}{2}dt\)
\(\left\{{}\begin{matrix}x=0\Rightarrow t=1\\x=\pi\Rightarrow t=1\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^1_1-\dfrac{1}{2}.t^2dt=0\) (hai cận bằng nhau thì tích phân bằng 0 khỏi tính dài dòng)
c. Đặt \(\left\{{}\begin{matrix}u=x\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=e^x\end{matrix}\right.\)
\(I=x.e^x|^1_0-\int\limits^1_0e^xdx=\left(x.e^x-e^x\right)|^1_0=1\)
a)
Ta có:
∫π20cos2xsin2xdx=12∫π20cos2x(1−cos2x)dx=12∫π20[cos2x−1+cos4x2]dx=14∫π20(2cos2x−cos4x−1)dx=14[sin2x−sin4x4−x]π20=−14.π2=−π8∫0π2cos2xsin2xdx=12∫0π2cos2x(1−cos2x)dx=12∫0π2[cos2x−1+cos4x2]dx=14∫0π2(2cos2x−cos4x−1)dx=14[sin2x−sin4x4−x]0π2=−14.π2=−π8
b)
Ta có: Xét 2x – 2-x ≥ 0 ⇔ x ≥ 0.
Ta tách thành tổng của hai tích phân:
∫1−1|2x−2−x|dx=−∫0−1(2x−2−x)dx+∫10(2x−2−x)dx=−(2xln2+2−xln2)∣∣0−1+(2xln2+2−xln2)∣∣10=1ln2∫−11|2x−2−x|dx=−∫−10(2x−2−x)dx+∫01(2x−2−x)dx=−(2xln2+2−xln2)|−10+(2xln2+2−xln2)|01=1ln2
c)
∫21(x+1)(x+2)(x+3)x2dx=∫21x3+6x2+11x+6x2dx=∫21(x+6+11x+6x2)dx=[x22+6x+11ln|x|−6x]∣∣21=(2+12+11ln2−3)−(12+6−6)=212+11ln2∫12(x+1)(x+2)(x+3)x2dx=∫12x3+6x2+11x+6x2dx=∫12(x+6+11x+6x2)dx=[x22+6x+11ln|x|−6x]|12=(2+12+11ln2−3)−(12+6−6)=212+11ln2
d)
∫201x2−2x−3dx=∫201(x+1)(x−3)dx=14∫20(1x−3−1x+1)dx=14[ln|x−3|−ln|x+1|]∣∣20=14[1−ln2−ln3]=14(1−ln6)∫021x2−2x−3dx=∫021(x+1)(x−3)dx=14∫02(1x−3−1x+1)dx=14[ln|x−3|−ln|x+1|]|02=14[1−ln2−ln3]=14(1−ln6)
e)
∫π20(sinx+cosx)2dx=∫π20(1+sin2x)dx=[x−cos2x2]∣∣π20=π2+1∫0π2(sinx+cosx)2dx=∫0π2(1+sin2x)dx=[x−cos2x2]|0π2=π2+1
g)
I=∫π0(x+sinx)2dx∫π0(x2+2xsinx+sin2x)dx=[x33]∣∣π0+2∫π0xsinxdx+12∫π0(1−cos2x)dxI=∫0π(x+sinx)2dx∫0π(x2+2xsinx+sin2x)dx=[x33]|0π+2∫0πxsinxdx+12∫0π(1−cos2x)dx
Tính :J=∫π0xsinxdxJ=∫0πxsinxdx
Đặt u = x ⇒ u’ = 1 và v’ = sinx ⇒ v = -cos x
Suy ra:
J=[−xcosx]∣∣π0+∫π0cosxdx=π+[sinx]∣∣π0=πJ=[−xcosx]|0π+∫0πcosxdx=π+[sinx]|0π=π
Do đó:
I=π33+2π+12[x−sin2x2]∣∣π30=π33+2π+π2=2π3+15π6
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 \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
\(\left\{{}\begin{matrix}u=x^2\\dv=cos2xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2xdx\\v=\dfrac{1}{2}sin2x\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{2}x^2sin2x|^{\pi}_0-\int\limits^{\pi}_0x.sin2xdx\)