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Data publicarii: 04 Noiembrie, 2015

EXERCITIUL 17

Suport teoretic:

Primitive,integrare prin parti,derivate. 

Enunt:

Sa se calculeze: 

I=\int{xsin^2x}dx.I=\int{xsin^2x}dx.  

Raspuns: 

I=\frac{x^2}{2}sin^2x+\frac{1}{4}x^2cos2x-\frac{1}{4}xsin2x-\frac{1}{8}cos2x+\mathcal{C}.I=\frac{x^2}{2}sin^2x+\frac{1}{4}x^2cos2x-\frac{1}{4}xsin2x-\frac{1}{8}cos2x+\mathcal{C}.  

Rezolvare:

I=\int{\big(\frac{x^2}{2}\big)}^{I=\int{\big(\frac{x^2}{2}\big)}^{'}sin^2xdx=\frac{x^2}{2}sin^2x-\int{\frac{x^2}{2}2sinxcosx}dx=

=\frac{x^2}{2}sin^2x-\frac{1}{2}\int{{x^2}sin2x}dx=\frac{x^2}{2}sin^2x-{\frac{1}{2}}\cdot{J}\cdot\;(1)=\frac{x^2}{2}sin^2x-\frac{1}{2}\int{{x^2}sin2x}dx=\frac{x^2}{2}sin^2x-{\frac{1}{2}}\cdot{J}\cdot\;(1)   

J=\int{{x^2}sin2x}dx=\int{x^2\big(\frac{-cos2x}{2}\big)^{J=\int{{x^2}sin2x}dx=\int{x^2\big(\frac{-cos2x}{2}\big)^{'}}dx=

=-\frac{x^2}{2}cos2x+\int{xcos2x}dx=-\frac{x^2}{2}cos2x+K\cdot\;(2)=-\frac{x^2}{2}cos2x+\int{xcos2x}dx=-\frac{x^2}{2}cos2x+K\cdot\;(2)  

K=\int{xcos2x}dx=\int{x\big(\frac{sin2x}{2}\big)^{K=\int{xcos2x}dx=\int{x\big(\frac{sin2x}{2}\big)^{'}}dx=\cdots= \frac{x}{2}sin2x+\frac{1}{4}cos2x+\mathcal{C}.\;(3)\frac{x}{2}sin2x+\frac{1}{4}cos2x+\mathcal{C}.\;(3)

Din (1), (2) si (3) rezulta: 

I=\frac{x^2}{2}sin^2x+\frac{1}{4}x^2cos2x-\frac{1}{4}xsin2x-\frac{1}{8}cos2x+\mathcal{C}.I=\frac{x^2}{2}sin^2x+\frac{1}{4}x^2cos2x-\frac{1}{4}xsin2x-\frac{1}{8}cos2x+\mathcal{C}.  

Observatie: 

Rezolvarea se poate realiza si astfel:

I=\int{xsin^2x}dx =I=\int{xsin^2x}dx = \int{x\cdot{\frac{1-cos2x}{2}}}dx\;etc.\int{x\cdot{\frac{1-cos2x}{2}}}dx\;etc.

Postat în: PRIMITIVE-liceu

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