[Toán 12] Chuyên đề: Nguyên hàm tích phân

[letrang3003]

tiếp tục
câu 2,

d,[TEX]I=\int \frac{dx}{(x^2+2x+5)^2}[/TEX]

 

[phamduyquoc0906]

[TEX]SAI\ \ \ \ \ x+1=2tgt\Rightarrow{dx=2(1+tg^2t)dt\Rightarrow{I= \frac{1}{8}\int{cos^2tdt=\frac{1}{16}\int{(1+cos2t)dt=\frac{1}{16}t+\frac{1}{32}sin2t+C=\frac{1}{16}arctg{\frac{x+1}{2}}+\frac{1}{8}\frac{x+1}{x^2+2x+5}+C[/TEX]

 

[bigbang195]

 

[bigbang195]

 

[phamduyquoc0906]

[TEX]1/ I=\int_{ln3}^{ln5}\frac{1}{e^x+2e^{-x}-3}dx=\int_{ln3}^{ln5}\frac{e^x}{e^{2x}-3e^x+2}dx[/TEX][TEX]\Rightarrow{t=e^x[/TEX]

[TEX]2/I=\int_0^{\frac{\pi}{12}}4sinx.sin2x.sin3xdx=\int_0^{\frac{\pi}{12}}2sin3x(cosx-cos3x)dx=\int_0^{\frac{\pi}{12}}(sin4x+sin2x-sin6x)dx[/TEX]

[TEX]3/I=\int\frac{x^2+1}{(x-1)^3(x+3)}dx=\int\frac{(x-1)^2+2x}{(x-1)^3(x+3)}dx=\int(\frac{\frac{1}{4}}{x-1}-\frac{\frac{5}{32}}{x+3}+-\frac{1}{32}\frac{3x^2-18x-1}{(x-1)^3})dx[/TEX]

Cái cuối đặt [TEX]t=x-1[/TEX]

[TEX]4/I=\int_0^{\frac{\pi}{2}}\frac{sin2x}{1+cos^2x}dx \Rightarrow{t=1+cos^2x[/TEX]

[TEX]5/I=\int_0^{\frac{\pi}{2}}\sqrt{1+3sinx}cosxdx \Rightarrow{t=\sqrt{1+3sinx}[/TEX]


[TEX]6/I=\int_0^{\frac{\pi}{2}}\sqrt{sinx-sin^3x}dx=\int_0^{\frac{\pi}{2}}\sqrt{sinx(1-sin^2x)}dx=\int_0^{\frac{\pi}{2}}\sqrt{sinx}cosxdx\Rightarrow{t=\sqrt{sinx}[/TEX]

[TEX]7/I=\int_0^1\sqrt{1+x^2}dx\Rightarrow{[/TEX] có ở trên rồi

 

[kimxakiem2507]

[TEX]1/\ \ \ \ I=\int_0^1\frac{x}{\sqrt{x+1}+\sqrt[3]{x+1}}dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [/TEX]

[TEX]2/\ \ \ \ \ \ I=\int_{\frac{1}{sqrt3}}^{\sqrt3}\frac{1}{1+x^2+x^4+x^6}dx\ \ \ \ \ \ \ \ \ \ [/TEX]


[TEX]3/\ \ \ \ I=\int_0^1e^{\ ^{2x-e^{x}}}dx[/TEX]

 

[letrang3003]

 

[letrang3003]

~X(

p/s: mấy anh cho em mấy bài dễ dễ đi ạ, mấy bài anh kimxakiem post khó quá ạ :-s

 

[letrang3003]

[TEX]2/\ \ \ \ \ \ I=\int_{\frac{1}{sqrt3}}^{\sqrt3}\frac{1}{1+x^2+x^4+x^6}dx\ \ \ \ \ \ \ \ \ \ [/TEX]

[TEX]1/\ \ \ \ I=\int_0^1\frac{x}{\sqrt{x+1}+\sqrt[3]{x+1}}dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [/TEX]

đặt

, đổi cận.....

khai triển nốt cái cuối là có kết quả

[TEX]3/\ \ \ \ I=\int_0^1e^{\ ^{2x-e^{x}}}dx[/TEX]

 

[vivietnam]

đặt x=-t\Rightarrowdx=-dt
\Rightarrow[TEX]I=\int\limits_{-1}^{1}\frac{sin^6t+cos^6t}{6^t+1}.6^tdt[/TEX]
\Rightarrow[TEX] 2I=\int\limits_{-1}^{1} (sin^6t+cos^6t)dt[/TEX]

đến đây chắc làm tiếp được
hạ bậc là ổn

 

[vivietnam]

[TEX]\int \frac{(3x-5)^{10}}{(x+2)^{12}}dx=\int \frac{1}{11}({\frac{3x-5}{x+2})^{10}.d(\frac{3x-5}{x+2})=...............[/TEX]
[TEX]\int \frac{(7x-1)^{99}}{(2x+1)^{101}}dx=\int \frac{1}{9}.(\frac{7x-1}{2x+1})^{99}d(\frac{3x-5}{2x+1})=....................[/TEX]

 

[hunggary]

tổng quát

1; [TEX]I = \int\limits_{}^{}cos^nx . sinnxdx = \frac{cos^nxcosnx}{2n} + \frac{1}{2} .\int_{}^{} co s ^{ n- 1} x sin ( n- 1) x dx[/TEX]

2; [TEX]I = \int\limits_{}^{}cos^nx . cosnxdx=\frac{cos^nx sin nx}{2n} + \frac{1}{2} .\int_{}^{} co s ^{ n- 1} x cos ( n- 1) x dx[/TEX]

không hiểu bài này
giải thích rõ ràng hơn đi
sao lại có cái cos(n-1)x với sin(n-1)x

Làm câu 1 thôi nha ..
Ta có :
[TEX]I_n = \int\limits_{0}^{\pi}cos^nx.[cos(n+1)x.cosx + sin(n+1)x.sinx]dx = \int\limits_{0}^{\pi}cos^{n+1}x.cos(n+1)x + \int\limits_{0}^{\pi}sin(n+1)x.cos^nx.sinx dx[/TEX] (1)
Đặt [TEX]u = sin(n+1)x \Rightarrow du = (n+1)cos(n+1)xdx[/TEX]
[TEX]v = \int\limits_{a}^{b}cos^nxsinxdx = - \int\limits_{a}^{b}cos^nxd(cosx) = - \frac{1}{n+1}.cos^{n+1}x[/TEX]
Theo CT tính tích phân từng phần thì :
[TEX]\int\limits_{0}^{\pi}sin(n+1)x.cos^nx.sinxdx = - \frac{1}{n+1}.cos^{n+1}x.sin(n+1)x |\frac{\pi}{0} + \int\limits_{0}^{\pi}cos^{n+1}.cos(n+1)xdx = \int\limits_{0}^{\pi}cos^{n+1}x.cos(n+1)xdx (2)[/TEX]
[TEX]I_n =2. \int\limits_{0}^{\pi}cos^{n+1}x.cos(n+1)xdx[/TEX]
[TEX]\Rightarrow I_n = 2I_{n+1}[/TEX] (3)
Áp dụng (3) ta có :
[TEX]I_0 = 2I_1 = 2^2.I_2 = .... = 2^n.I_n[/TEX]
Vậy : [TEX]I_n = \frac{1}{2^n}.I_0[/TEX] (4)
Do [TEX]I_0 = \int\limits_{0}^{\pi}dx = \pi[/TEX]
Vậy [TEX]I_n = \frac{\pi}{2^n}[/TEX]

 

[kimxakiem2507]

[TEX]I=\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{1}{1+x^2+x^{4}+x^{6}}dx[/TEX][TEX](t=\frac{1}{x})[/TEX]
[TEX]\Rightarrow{I_{*}=\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{1}{1+x^2+x^{2010}+x^{2012}}dx[/TEX]

[TEX]I= \int_{\frac{1}{a}}^{a}\frac{f(x)}{(1+x^2)(1+x^{n})}dx\ \ \ \ \ \ f(x)=f(\frac{1}{x})[/TEX]

[TEX]t=\frac{1}{x}\Rightarrow{dt=-\frac{1}{t^2}dt[/TEX]
[TEX]\Rightarrow{I=\int_{\frac{1}{a}}^{a}\frac{\frac{1}{t^2}.f(\frac{1}{t})}{(1+\frac{1}{t^2})(1+\frac{1}{t^{n}})}dt=\int_{\frac{1}{a}}^{a}\frac{t^{n}.f(t)}{(1+t^2)(1+t^{n})}dt==\int_{\frac{1}{a}}^{a}\frac{x^{n}.f(x)}{(1+x^2)(1+x^{n})}dx=I[/TEX]

[TEX]\Rightarrow{I=\frac{1}{2} \int_{\frac{1}{a}}^{a}\frac{f(x)}{1+x^2}dx[/TEX]

[TEX]I=\int_{-m}^{m}\frac{f(x)}{1+a^{g(x)}}dx[/TEX][TEX]\ \ \ \ \ \ \ \left{f(x)=f(-x)\\g(x)=-g(-x)\\a>0[/TEX]

Đặt [TEX]x=-t\Rightarrow{dx=-dt[/TEX]

[TEX]\Rightarrow{I=-\int_{m}^{-m}\frac{f(-t)}{1+a^{g(-t)}}dt=\int_{-m}^{m}\frac{f(t)}{1+a^{-g(t)}}}dt=\int_{-m}^{m}\frac{a^{g(t)}.f(t)}{1+a^{g(t)}}dt=\int_{-m}^{m}\frac{a^{g(x)}.f(x)}{1+a^{g(x)}}dx[/TEX]

[TEX]I+I=\int_{-m}^{m}\frac{f(x)}{1+a^{g(x)}}dx+\int_{-m}^{m}\frac{a^{g(x)}.f(x)}{1+a^{g(x)}}dx=\int_{-m}^{m} f(x)dx[/TEX]

[TEX]\Leftrightarrow{I=\frac{1}{2}\int_{-m}^{m} f(x)dx=\int_{0}^{m} f(x)dx[/TEX]

 

[phamduyquoc0906]

[TEX]I=\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{1}{1+x^2+x^{4}+x^{6}}dx[/TEX][TEX](t=\frac{1}{x})[/TEX]
[TEX]\Rightarrow{I_{*}=\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{1}{1+x^2+x^{2010}+x^{2012}}dx[/TEX]

[TEX]t=\frac{1}{x}\Rightarrow{I=\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{\frac{1}{t^2}}{1+\frac{1}{t^2}+ \frac{1}{t^4}+\frac{1}{t^6}}dt=\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{x^4}{(1+x^2)(1+x^4)}dx[/TEX]

[TEX]\Rightarrow{2I=\int_{ \frac{1}{\sqrt3}}^{\sqrt3}\frac{1}{1+x^2}dx=arctgx\|_{\frac{1}{\sqrt3}}^{\sqrt3}=\frac{\pi}{6} \Leftrightarrow{I=\frac{\pi}{12}[/TEX]


[TEX]t=\frac{1}{x}\Rightarrow{I_{*}=\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{\frac{1}{t^2}}{1+\frac{1}{t^2}+ \frac{1}{t^{2010}}+\frac{1}{t^{2012}}}dt= \int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{x^{2010}}{(1+x^2)(1+x^{2010})}dx[/TEX]

[TEX]\Rightarrow{2I_{*}=\int_{ \frac{1}{\sqrt3}}^{\sqrt3}\frac{1}{1+x^2}dx=arctgx\|_{\frac{1}{\sqrt3}}^{\sqrt3}=\frac{\pi}{6} \Leftrightarrow{I_{*}=\frac{\pi}{12}[/TEX]

 

[keosuabeo_93]

1.[tex]\int\limits_{0}^{1}x.ln(x^2+x+1)dx[/tex]

2.[tex]\int\limits_{0}^{\frac{\pi}{2}}sin^5xdx[/tex]

 

[phamduyquoc0906]

1.[tex]\int\limits_{0}^{1}x.ln(x^2+x+1)dx[/tex]

2.[tex]\int\limits_{0}^{\frac{\pi}{2}}sin^5xdx[/tex]

[TEX]1/I=\frac{1}{2}\int{ln(x^2+x+1)d(x^2)=\frac{1}{2}x^2ln(x^2+x+1)-\frac{1}{2}\int{\frac{(2x+1)x^2}{x^2+x+1}dx=\frac{1}{2}x^2ln(x^2+x+1)-\frac{1}{2}\int(2x-1-\frac{1}{2}\frac{2x+1}{x^2+x+1}+\frac{3}{2}\frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}})dx[/TEX]

[TEX]2/I=\int{sin^5xdx=-\int{(1-cos^2x)^2d(cosx)=\int(2cos^2x-cos^4x-1)d(cosx)=\frac{2}{3}cos^3x-\frac{1}{5}cos^5x-cosx+C[/TEX]

 

[letrang3003]

tách :

hơi dài :-s


bài dưới :

 

[phamduyquoc0906]

[tex] \int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{1}{1+x^2+x^4+x^6}dx=\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{x^4}{(1+x^2)(1+x^4)}dx[/tex] ????????
bạn giải thích giùm mình cái mình nghĩ cái này sai òy

[TEX]t=\frac{1}{x}\Rightarrow{dx=-\frac{1}{t^2}dt[/TEX]

[TEX]I=\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{\frac{1}{t^2}}{1+\frac{1}{t^2}+\frac{1}{t^4}+\frac{1}{t^6}}dt=\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{t^4}{1+t^2+t^4+t^6}dt=\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{t^4}{(1+t^2)(1+t^4)}dt=\int_{\frac{1}{\sqrt3}}^{ \sqrt3}\frac{x^4}{(1+x^2)(1+x^4)}dx[/TEX]


[TEX]1+x^2+x^4+x^6=(1+x^2)(1+x^4)[/TEX]

 

[letrang3003]

tiếp tục
câu 2,
a, [TEX]I=\int arctanxdx[/TEX]
b,[TEX]I=\int \frac{xdx}{\sqrt{x^2+x+2}}[/TEX]
c,[TEX]I=\int x.\sqrt{-x^2+3x-2}dx[/TEX]
d,[TEX]I=\int \frac{dx}{(x^2+2x+5)^2}[/TEX]
e,[TEX]I=\int sin^{n-1}x.sin(n+1)xdx[/TEX]
f,[TEX]I=\int e^{-2x}cos3xdx[/TEX]
g,[TEX]I= \int arcsin^2xdx[/TEX]

a/

b/

đặt

c/

d/ đã làm

e/

mà

thế vào (1)

 

[nhockthongay_girlkute]

[TEX]\int\limits_0^{\frac{\pi}{2}}e^{sinx}Cos^3 xd x[/TEX]