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Data publicarii: 26 Octombrie, 2013

EXERCITIUL 6

Suport teoretic:

Primitive,integrare prin parti,derivate,formule de recurenta.

Enunt:

Fie integrala:

I_n=\int{arcsin^{n}xdx,\;x\in{(-1;+1)},\;n\in{N},\;n\ge{1}}.I_n=\int{arcsin^{n}xdx,\;x\in{(-1;+1)},\;n\in{N},\;n\ge{1}}.

Sa se calculeze I1 , I2 si apoi sa se stabileasca o relatie de recurenta pentru In, n > 2.

Raspuns:

I_1=x\cdot{arcsinx}+\sqrt{1-x^2}+\mathcal{C}.I_1=x\cdot{arcsinx}+\sqrt{1-x^2}+\mathcal{C}.

I_2=x\cdot{arcsin^2x}+2{\sqrt{1-x^2}}\cdot{arcsinx}-2x+\mathcal{C}.I_2=x\cdot{arcsin^2x}+2{\sqrt{1-x^2}}\cdot{arcsinx}-2x+\mathcal{C}.

I_n=x\cdot{arcsin^nx}+n{\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}-n(n-1)I_{n-2}.I_n=x\cdot{arcsin^nx}+n{\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}-n(n-1)I_{n-2}.

Rezolvare:

I_1=\int{arcsinxdx}=\int{{xI_1=\int{arcsinxdx}=\int{{x'}\cdot{arcsinxdx}}=x\cdot{arcsinx}-\int{x\cdot{(arcsinx)'dx}}= =x\cdot{arcsinx}-\int{{x}\cdot{\frac{1}{\sqrt{1-x^2}}}}dx=x\cdot{arcsinx}+\int{\frac{-2x}{2\sqrt{1-x^2}}}dx==x\cdot{arcsinx}-\int{{x}\cdot{\frac{1}{\sqrt{1-x^2}}}}dx=x\cdot{arcsinx}+\int{\frac{-2x}{2\sqrt{1-x^2}}}dx= x\cdot{arcsinx}+\int{\frac{(1-x^2)x\cdot{arcsinx}+\int{\frac{(1-x^2)'}{2\sqrt{1-x^2}}}dx=

=x\cdot{arcsinx}+\int{(\sqrt{1-x^2})=x\cdot{arcsinx}+\int{(\sqrt{1-x^2})'}dx= x\cdot{arcsinx}+\sqrt{1-x^2}+\mathcal{C}.x\cdot{arcsinx}+\sqrt{1-x^2}+\mathcal{C}.

I_2=\int{arcsin^2xdx}=\int{{xI_2=\int{arcsin^2xdx}=\int{{x'}\cdot{arcsin^2xdx}}=x\cdot{arcsin^2x}-\int{x\cdot{(arcsin^2x)'dx}}= =x\cdot{arcsin^2x}-\int{{x}\cdot{\frac{2arcsinx}{\sqrt{1-x^2}}}}dx=x\cdot{arcsin^2x}+2\int{\frac{-2x}{2\sqrt{1-x^2}}\cdot{arcsinx}}dx==x\cdot{arcsin^2x}-\int{{x}\cdot{\frac{2arcsinx}{\sqrt{1-x^2}}}}dx=x\cdot{arcsin^2x}+2\int{\frac{-2x}{2\sqrt{1-x^2}}\cdot{arcsinx}}dx= =x\cdot{arcsin^2x}+2\int{\frac{(1-x^2)=x\cdot{arcsin^2x}+2\int{\frac{(1-x^2)'}{2\sqrt{1-x^2}}\cdot{arcsinx}}dx=

=x\cdot{arcsin^2x}+2\int{(\sqrt{1-x^2})=x\cdot{arcsin^2x}+2\int{(\sqrt{1-x^2})'}\cdot{arcsinx}dx=

=x\cdot{arcsin^2x}+2{\sqrt{1-x^2}}\cdot{arcsinx}-2\int{{\sqrt{1-x^2}}\cdot{(arcsinx)=x\cdot{arcsin^2x}+2{\sqrt{1-x^2}}\cdot{arcsinx}-2\int{{\sqrt{1-x^2}}\cdot{(arcsinx)'}dx}=

=x\cdot{arcsin^2x}+2{\sqrt{1-x^2}}\cdot{arcsinx}-2\int{{\sqrt{1-x^2}}\cdot{\frac{1}{\sqrt{1-x^2}}dx}}==x\cdot{arcsin^2x}+2{\sqrt{1-x^2}}\cdot{arcsinx}-2\int{{\sqrt{1-x^2}}\cdot{\frac{1}{\sqrt{1-x^2}}dx}}=

=x\cdot{arcsin^2x}+2{\sqrt{1-x^2}}\cdot{arcsinx}-2x+\mathcal{C}.=x\cdot{arcsin^2x}+2{\sqrt{1-x^2}}\cdot{arcsinx}-2x+\mathcal{C}.

I_n=\int{arcsin^{n}xdx}=I_n=\int{arcsin^{n}xdx}= \int{{x\int{{x'}\cdot{arcsin^nxdx}}=

=x\cdot{arcsin^nx}-\int{x\cdot{(arcsin^nx)=x\cdot{arcsin^nx}-\int{x\cdot{(arcsin^nx)'dx}}=

=x\cdot{arcsin^nx}-\int{{x}\cdot{\frac{narcsin^{n-1}x}{\sqrt{1-x^2}}}}dx==x\cdot{arcsin^nx}-\int{{x}\cdot{\frac{narcsin^{n-1}x}{\sqrt{1-x^2}}}}dx=

=x\cdot{arcsin^nx}+n\int{\frac{-2x}{2\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}}dx==x\cdot{arcsin^nx}+n\int{\frac{-2x}{2\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}}dx=

=x\cdot{arcsin^nx}+n\int{\frac{(1-x^2)=x\cdot{arcsin^nx}+n\int{\frac{(1-x^2)'}{2\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}}dx=

=x\cdot{arcsin^nx}+n\int{(\sqrt{1-x^2})=x\cdot{arcsin^nx}+n\int{(\sqrt{1-x^2})'}\cdot{arcsin^{n-1}x}dx=

=x\cdot{arcsin^nx}+n{\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}-n\int{{\sqrt{1-x^2}}\cdot{(arcsin^{n-1}x)=x\cdot{arcsin^nx}+n{\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}-n\int{{\sqrt{1-x^2}}\cdot{(arcsin^{n-1}x)'}dx}=

=x\cdot{arcsin^nx}+n{\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}-n\int{{\sqrt{1-x^2}}\cdot{\frac{(n-1)\cdot{arcsin^{n-2}x}}{\sqrt{1-x^2}}}dx}==x\cdot{arcsin^nx}+n{\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}-n\int{{\sqrt{1-x^2}}\cdot{\frac{(n-1)\cdot{arcsin^{n-2}x}}{\sqrt{1-x^2}}}dx}=

=x\cdot{arcsin^nx}+n{\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}-n(n-1)\int{arcsin^{n-2}xdx}==x\cdot{arcsin^nx}+n{\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}-n(n-1)\int{arcsin^{n-2}xdx}=

=x\cdot{arcsin^nx}+n{\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}-n(n-1)I_{n-2}.=x\cdot{arcsin^nx}+n{\sqrt{1-x^2}}\cdot{arcsin^{n-1}x}-n(n-1)I_{n-2}.

Postat în: PRIMITIVE-liceu

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