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Mathematics 9 Online
OpenStudy (anonymous):

Integral SOS!

OpenStudy (anonymous):

\[\int\limits \sin \sqrt{(2x+1)}dx\]

OpenStudy (anonymous):

it's the sin of this root

rvc (rvc):

\[\sqrt{2x+1}=t\]

OpenStudy (anonymous):

yeah I did that, and from there I got stuck....

OpenStudy (michele_laino):

hint: try this substitution: \[2x + 1 = {t^2}\]

OpenStudy (anonymous):

oh no, why?

OpenStudy (anonymous):

I got almost the answer, beside 2 as the coefficient

rvc (rvc):

okay\[\sqrt{2x+1}=t ---->(2x+1)= t^{2}--->2dx/dt=2t\]

rvc (rvc):

\[dx=tdt\]

OpenStudy (michele_laino):

we can rewrite your integral as below: \[\int {t\sin t\;dt} \]

rvc (rvc):

\[\int\limits tsintdt--> now \ use \\ integration \ by\ parts\]

rvc (rvc):

LIATE<---follow this order

OpenStudy (anonymous):

I didn't get why 2dx/dt=2t

rvc (rvc):

oh wait

rvc (rvc):

|dw:1427364321419:dw|

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