Определен интеграл

[a40o0]

Малко помощ с тези задачи

dx.doc

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[Mark]

[tex]\int_{0}^{\pi} xcos^2xdx=\frac{\pi}{2}\int_{0}^{\pi} cos^2xdx=[/tex][tex]\frac{\pi}{2}(\frac{x}{2}+\frac{sin2x}{4})|_{0}^{\pi}=\frac{\pi^2}{4}[/tex]
[tex]\int_{3}^{6} \frac{\sqrt{x^2-9}}{x^2}dx , x=\frac{3}{\sqrt{1-u}}[/tex]
[tex]\int \frac{\sqrt{u}}{1-u}du=\int\frac{t^2}{1-t^2}dt[/tex][tex]=-\int dt+\frac{1}{2}\int \frac{1}{1+t}dt+\frac{1}{2}\int \frac{1}{1-t}dt=[/tex][tex]\frac{1}{2}ln|\frac{1+t}{1-t}|-t[/tex]
[tex]\int_{3}^{6} \frac{\sqrt{x^2-9}}{x^2}dx[/tex][tex]=\frac{1}{2}ln|\frac{1+t}{1-t}|-t|_{0}^{\frac{\sqrt{3}}{2}}=ln(2+\sqrt{3})-\frac{\sqrt{3}}{2}[/tex]

[stflyfisher]

[tex]\int_{3}^{6} \frac{\sqrt{x^2-9}}{x^2}dx[/tex]

Интеграли от вида:

[tex]\int R(x,\sqrt{x^2-a^2})dx, R[/tex]-рационална ф-ция на [tex]x[/tex] и [tex]\sqrt{x^2-a^2}[/tex]

се решават стандартно със субституцията:

[tex]x=\frac{a}{sint}[/tex] или [tex]x=\frac{a}{cos t}[/tex]

Този интеграл може да се реши и пo този начин:

[tex]\int_{3}^{6} \frac{\sqrt{x^2-9}}{x^2}dx=\int_{3}^{6} \frac{\sqrt{x^2-3^2}}{x^2}dx[/tex]

[tex]x=\frac{3}{sint}=>dx=-\frac{3}{sin^2t}costdt=>t=arcsin\frac{3}{x}[/tex]

[tex]x=3=>t=\frac{ \pi}{2}[/tex]

[tex]x=6=>t=\frac{ \pi}{6}[/tex]

[tex]\int_{3}^{6} \frac{\sqrt{x^2-3^2}}{x^2}dx=\int_{\frac{\pi}{2}}^{\frac{\pi}{6} }\frac{\sqrt{\frac{9}{sin^2t}-9}}{\frac{9}{sin^2t}}.(-\frac{3}{sin^2t}.cost)dt=...[/tex]

[tex]...=-\int_{\frac{\pi}{2}}^{\frac{\pi}{6} }\sqrt{\frac{1-sin^2t}{sin^2t}}.costdt=-\int_{\frac{\pi}{2}}^{\frac{\pi}{6} }\frac{\sqrt{1-sin^2t}}{\sqrt{sin^2t}}.cosdtt=[/tex]

[tex]=-\int_{\frac{\pi}{2}}^{\frac{\pi}{6} }\frac{|cost|}{|sint|}.cosdtt=-\int_{\frac{\pi}{2}}^{\frac{\pi}{6} }\frac{cost}{sint}.costdt=[/tex]

[tex]=-\int_{\frac{\pi}{2}}^{\frac{\pi}{6} }\frac{cos^2t}{sint}dt=-\int_{\frac{\pi}{2}}^{\frac{\pi}{6} }\frac{1-sin^2t}{sint}dt=[/tex]

[tex]=\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }\frac{1-sin^2t}{sint}dt[/tex]

[tex]=\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }\frac{1}{sint}dt-\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }sintdt=A-B[/tex]

[tex]A=\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }\frac{1}{sint}dt[/tex] се решава:

1.) с помощта на универсалната субституция:

[tex]tg\frac{t}{2}=z, dt=\frac{2}{1+z^2}dz, sint=\frac{2z}{1+z^2}dz[/tex]

[tex]t=\frac{\pi}{6}=>z=tg{\frac{\pi}{12}[/tex]

[tex]t=\frac{\pi}{2}=>z=1[/tex]

[tex]A=\int_{\frac{\pi}{2}}^{\frac{\pi}{6} }\frac{1}{sint}dt=\int_{tg(\frac{\pi}{12})}^{1 }\frac{1}{\frac{2z}{1+z^2}}.\frac{2}{1+z^2}dt=...=[/tex]

[tex]=ln|{z}|_{tg(\frac{\pi}{12})}^1=ln|1|-ln|tg(\frac{\pi}{12})|[/tex]

[tex]tg \frac{\alpha }{2}=\frac{sin\alpha }{1+cos\alpha }=>[/tex]

[tex]tg(\frac{\pi}{12})=tg(\frac{\frac{\pi}{6}}{2})=\frac{sin( \frac{\pi}{6})}{1+cos (\frac{\pi}{6})}=[/tex]

[tex]=\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}=2-\sqrt{3}=>[/tex]

[tex]A=0-ln|2-\sqrt{3}|=-ln(2-\sqrt{3})[/tex]

или се решава:

[stflyfisher]

2.)[tex]A=\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }\frac{1}{sint}dt=\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }\frac{1}{2sin(\frac{t}{2})cos(\frac{t}{2})}dt=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }\frac{1}{\frac{sin(\frac{t}{2})}{cos(\frac{t}{2})}cos^2(\frac{t}{2})}dt=[/tex]

[tex]=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }\frac{1}{tg(\frac{t}{2})cos^2(\frac{t}{2})}dt=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }\frac{1}{tg(\frac{t}{2})}.\frac{1}{cos^2(\frac{t}{2})}dt=[/tex]

[tex]=\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }\frac{1}{tg(\frac{t}{2})}.\frac{1}{cos^2(\frac{t}{2})}d(\frac{t}{2})=\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }\frac{1}{tg(\frac{t}{2})}d tg(\frac{t}{2})=[/tex]

[tex]=ln|tg(\frac{t}{2})||_{\frac{\pi}{6}}^{\frac{\pi}{2}}=ln|tg(\frac{\pi}{4})|-ln|tg(\frac{\pi}{12})=ln|1|-ln|2-\sqrt{3}|=ln 1-ln(2-\sqrt{3})=-ln(2-\sqrt{3})[/tex]

[tex]B=\int_{\frac{\pi}{6}}^{\frac{\pi}{2} }sintdt=-cost|_{\frac{\pi}{6}}^{\frac{\pi}{2} }=-[cos(\frac{\pi}{2})-cos(\frac{\pi}{6})]=-[0-\frac{\sqrt{3}}{2}]=\frac{\sqrt{3}}{2}[/tex]

Окончателно:

[tex]\int_{3}^{6} \frac{\sqrt{x^2-9}}{x^2}dx=A-B=-ln(2-\sqrt{3})-\frac{\sqrt{3}}{2}[/tex],

което съвпада с отговора даден от Mark:

[tex]\int_{3}^{6} \frac{\sqrt{x^2-9}}{x^2}dx=ln(2+\sqrt{3})-\frac{\sqrt{3}}{2}[/tex]