Can someone show me how to solve this step by step using U substitution
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OpenStudy (anonymous):
OpenStudy (anonymous):
u=x^2 +5
OpenStudy (anonymous):
derrive u
OpenStudy (anonymous):
2x=du/dx
OpenStudy (anonymous):
solve for dx
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OpenStudy (anonymous):
2x=du times dx
OpenStudy (anonymous):
i mean du=2xdx
OpenStudy (anonymous):
then solve for dx
OpenStudy (anonymous):
dx=du/2x
OpenStudy (anonymous):
plug u and dx back into the integral
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OpenStudy (anonymous):
integral xsin(u) dx
OpenStudy (anonymous):
plug in your info
OpenStudy (anonymous):
for dx
OpenStudy (anonymous):
integral xsin(u)du/2x
OpenStudy (anonymous):
sorry hold
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OpenStudy (anonymous):
u can take out a 1/ 2x to the front
OpenStudy (anonymous):
1/2x integral xsin (u)
OpenStudy (anonymous):
now integrate
OpenStudy (anonymous):
and you fet 1/2x cosx(u) +c
OpenStudy (anonymous):
plug u back into (u)
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OpenStudy (anonymous):
not sure if this is right though
zepdrix (zepdrix):
Hmm I don't understand why there is x and u in your integral alexis D:
I'm just a lil confused.
Lemme post some notes if it will help.
\[\Large\bf\sf \int\sin(\color{orangered}{x^2+5})(\color{royalblue}{x\;dx})\]
\[\Large\bf\sf \color{orangered}{u=x^2+5}, \qquad\qquad\qquad \color{royalblue}{\frac{1}{2}du=x\;dx}\]
Becomes:\[\Large\bf\sf \int\limits\sin(\color{orangered}{u})(\color{royalblue}{\frac{1}{2}\;du})\quad=\quad \frac{1}{2}\int\limits \sin u \;du\]