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Mathematics 14 Online
OpenStudy (anonymous):

can someone help me figure out this integral, i will write it out

OpenStudy (anonymous):

\[\int\limits_{}^{}(x+sinx)^3+cosx(x+sinx)^3 dx\]

OpenStudy (anonymous):

u substitution

OpenStudy (anonymous):

u=x+sinx

OpenStudy (anonymous):

Let u=(x + sinx)

OpenStudy (anonymous):

okay so then du=1+cosx dx

OpenStudy (anonymous):

right

hero (hero):

But first, use distributive property to re-write it like this: \[\int\limits\limits_{}^{}(x+sinx)^3(1+cosx) dx\]

hero (hero):

That ought to make it easier

OpenStudy (anonymous):

how though if you distriubte cosx into (x+sinx), how do you get 1+cosx

hero (hero):

You distribute (x+sinx)^3 out of 1 + cosx not into it

hero (hero):

distributive property is obviously ab+ ac = a(b+c)....use your imagination here

OpenStudy (anonymous):

i still cant see that

hero (hero):

twiddla

OpenStudy (anonymous):

okay where

hero (hero):

http://www.twiddla.com/solved

OpenStudy (anonymous):

oh i see it, how you did it

OpenStudy (anonymous):

now i let u-1+sinx right

hero (hero):

wait, no u still equals x + sin x

OpenStudy (anonymous):

so will the answer be (1/4)(x+sinx)^4

hero (hero):

yes

hero (hero):

+ C

OpenStudy (anonymous):

thanks man

hero (hero):

\[\frac{(x+\sin(x))^4}{4}+c\]

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