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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
Lời giải:
Ta có : \(10=\int ^{3}_{1}f(2x)dx=\frac{1}{2}\int ^{3}_{1}f(2x)d(2x)=\frac{1}{2}\int ^{6}_{2}f(x)dx\)
\(\Rightarrow \int ^{6}_{2}f(x)d(x)=20\)
Mà \(\int ^{2}_{0}f(x)dx=-5\Rightarrow \int ^{6}_{0}f(x)dx=15\)
Do đó mà \(\int ^{2}_{0}f(3x)dx=\frac{1}{3}\int ^{2}_{0}f(3x)d(3x)=\frac{1}{3}\int ^{6}_{0}f(x)dx=5\)
Câu 1)
Ta có \(I=\int ^{1}_{0}\frac{dx}{\sqrt{3+2x-x^2}}=\int ^{1}_{0}\frac{dx}{4-(x-1)^2}\).
Đặt \(x-1=2\cos t\Rightarrow \sqrt{4-(x-1)^2}=\sqrt{4-4\cos^2t}=2|\sin t|\)
Khi đó:
\(I=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{d(2\cos t+1)}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{2\sin tdt}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}dt=\left.\begin{matrix} \frac{2\pi}{3}\\ \frac{\pi}{2}\end{matrix}\right|t=\frac{\pi}{6}\)
Câu 3)
\(K=\int ^{3}_{2}\ln (x^3-3x+2)dx=\int ^{3}_{2}\ln [(x+2)(x-1)^2]dx\)
\(=\int ^{3}_{2}\ln (x+2)d(x+2)+2\int ^{3}_{2}\ln (x-1)d(x-1)\)
Xét \(\int \ln tdt\): Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln t dt=t\ln t-t\)
\(\Rightarrow K=\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x+2)[\ln (x+2)-1]+2\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x-1)[\ln (x-1)-1]\)
\(=5\ln 5-4\ln 4-1+4\ln 2-2=5\ln 5-4\ln 2-3\)
Bài 2)
\(J=\int ^{1}_{0}x\ln (2x+1)dx\). Đặt \(\left\{\begin{matrix} u=\ln (2x+1)\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2dx}{2x+1}\\ v=\frac{x^2}{2}\end{matrix}\right.\)
Khi đó:
\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)
\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)
\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)
Chọn D
Xét \(I=\int\limits^2_0f\left(3x\right)dx\)
Đặt \(3x=t\Rightarrow dx=\dfrac{1}{3}dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=2\Rightarrow t=6\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^6_0f\left(t\right).\dfrac{1}{3}dt=\dfrac{1}{3}\int\limits^6_0f\left(t\right)dt=\dfrac{1}{3}\int\limits^6_0f\left(x\right)dx=\dfrac{1}{3}.12=4\)