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29 tháng 11 2019
https://i.imgur.com/Pe6vPSJ.jpg
AH
Akai Haruma
Giáo viên
6 tháng 3 2017

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 \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln t dt=t\ln t-t\)

\(\Rightarrow K=\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x+2)[\ln (x+2)-1]+2\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x-1)[\ln (x-1)-1]\)

\(=5\ln 5-4\ln 4-1+4\ln 2-2=5\ln 5-4\ln 2-3\)

AH
Akai Haruma
Giáo viên
6 tháng 3 2017

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:

\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)

\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)

\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)

1 tháng 4 2017

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∫0π2cos⁡2xsin2xdx=12∫0π2cos⁡2x(1−cos⁡2x)dx=12∫0π2[cos⁡2x−1+cos⁡4x2]dx=14∫0π2(2cos⁡2x−cos⁡4x−1)dx=14[sin⁡2x−sin⁡4x4−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=−(2xln⁡2+2−xln⁡2)|−10+(2xln⁡2+2−xln⁡2)|01=1ln⁡2

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+11ln⁡2−3)−(12+6−6)=212+11ln⁡2

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−ln⁡2−ln⁡3]=14(1−ln⁡6)

e)

∫π20(sinx+cosx)2dx=∫π20(1+sin2x)dx=[x−cos2x2]∣∣π20=π2+1∫0π2(sinx+cosx)2dx=∫0π2(1+sin⁡2x)dx=[x−cos⁡2x2]|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+2xsin⁡x+sin2x)dx=[x33]|0π+2∫0πxsin⁡xdx+12∫0π(1−cos⁡2x)dx

Tính :J=∫π0xsinxdxJ=∫0πxsin⁡xdx

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

4 tháng 4 2016

\(=\frac{1}{2}\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\ln\left(\tan x\right)d\left[\ln\left(\tan x\right)\right]=\frac{1}{4}\left[\ln^2\left(\tan x\right)\right]|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{4}\left(\ln^2\sqrt{3}-0\right)=\frac{1}{16}\ln^23\)

4 tháng 4 2016

Đặt \(t=\tan x\Rightarrow\begin{cases}dt=\frac{dt}{\cos^2}=\left(1+t^2\right)dx\rightarrow dx=\frac{dt}{1+t^2}\\x=\frac{\pi}{4}\rightarrow t=1;x=\frac{\pi}{3}\rightarrow t=\sqrt{3}\end{cases}\)

Khi đó : \(I=\int\limits^{\sqrt{3}}_1\frac{\ln t}{\frac{2t}{1+t^2}}.\frac{dt}{1+t^2}=\frac{1}{2}\int\limits^{\sqrt{3}}_1\frac{\ln t}{t}dt=\frac{1}{2}J\left(1\right)\)

\(J=\int\limits^{\sqrt{3}}_1\frac{\ln t}{t}dt=\int\limits^{\sqrt{3}}_1\ln.d\left(\ln t\right)=\frac{1}{2}\ln^2t|^{\sqrt{3}}_1=\frac{1}{2}\left(\ln^2\sqrt{3}-0\right)=\frac{1}{8}\ln^23\)

Thay vào (1) ta có : \(I=\frac{1}{16}\ln^23\)

27 tháng 4 2017

Hỏi đáp Toán

11 tháng 4 2016

\(I=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\cos2x+3\cos x+2}dx=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{2\cos^2x+3\cos x+1}dx\)

Đặt \(\cos x=t\Rightarrow dt=-\sin dx\)

Với \(x=0\Rightarrow t=1\)

Với \(x=\frac{\pi}{2}\Rightarrow t=0\)

\(I=\int\limits^1_0\frac{dt}{2t^2+3t+1}=\int\limits^1_0\frac{dt}{\left(2t+1\right)\left(t+1\right)}=2\int\limits^1_0\left(\frac{1}{2t+1}+\frac{1}{2t+1}\right)dt\)

  \(=\left(\ln\frac{2t+1}{2t+1}\right)|^1_0=\ln\frac{3}{2}\)