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OpenStudy (anonymous):
\[\int\limits_{?}^{?}(xcos(x^3))^2dx\]
OpenStudy (anonymous):
heres ? in the question!
OpenStudy (anonymous):
multiply it out first, will turn your integral into one that can be solved with a substitution.
OpenStudy (anonymous):
there is a part missing.
is it equal to something?
OpenStudy (anonymous):
@Dodo1 ?
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OpenStudy (anonymous):
No not equal!
OpenStudy (anonymous):
@electrokid
OpenStudy (anonymous):
It seems that the question is incomplete!!!
OpenStudy (anonymous):
let me write it again
OpenStudy (anonymous):
ok
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OpenStudy (anonymous):
if you can post a picture of the problem that would be nice too
OpenStudy (anonymous):
\[\int\limits_{}^{}(xcos(x^3))^2)dx \]
OpenStudy (anonymous):
sorry I wont be able to post photos my phone is not working but this is the question which is written in the paper.
OpenStudy (anonymous):
in the original problem you have "?"symbols at the integration limits
OpenStudy (anonymous):
it also says compute te following intergrals
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OpenStudy (anonymous):
Sorry theres no ?!
OpenStudy (anonymous):
∫(xcos(x^3))2)dx=∫1/3x^3(cos(x^3))^2)dx=∫1/3(cos(x^3))2)dx^3=∫1/3(cost)^2dt
=∫1/3(cos2t/2+1/2)dt
u should be able to do the rest
OpenStudy (anonymous):
thats what confused the people here...
so\[
\int[x\cos(x^3)]^2dx=\int x^2\cos^2(x^3)dx
\]
solve by substitution
\[u=x^3\implies du=3x^3dx\\
{du\over3}=x^2dx\]
so, we get
\[\int\cos^2(u){du\over3}={1\over3}\int\cos^2(u)du\]
then we have the double angle rule:
\[\cos^2\theta={\cos2\theta-1\over2}\]
OpenStudy (anonymous):
so, it reduces to
\[{1\over3}\int\left(\cos(2u)-1\over2\right)du\\
\quad={1\over6}\int[\cos(2u)-1]du\\
\quad={1\over6}\left[????\right]
\]
OpenStudy (anonymous):
you can solve from here :)
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OpenStudy (anonymous):
once you get the integration, replace "u" back by the substituion in "x"
OpenStudy (anonymous):
cos2*x^3??
OpenStudy (anonymous):
du =1/3?/
OpenStudy (anonymous):
dont put in "x" yet..
lets solve the integral in terms of "u"
OpenStudy (anonymous):
u(cos2)-(1)?
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OpenStudy (anonymous):
\[
{1\over6}\int\cos(2u)du-{1\over6}\int du
\]
OpenStudy (anonymous):
I separated them
OpenStudy (anonymous):
what is the integral of cos(2u)?
OpenStudy (anonymous):
i am not 100 % the defition of integral...
OpenStudy (anonymous):
is it derivative?
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OpenStudy (anonymous):
it is opposite of derivatives
"derivative of what is cos(2u)?"
OpenStudy (anonymous):
cosu^2?
OpenStudy (anonymous):
check.
do not guess
take the derivative of what you just said and see if you get cos(2u)
OpenStudy (anonymous):
i am not guessing. i actually thought it was the answer.
OpenStudy (anonymous):
hmm
whose derivative is cos(u)??
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OpenStudy (anonymous):
2cosu?
OpenStudy (anonymous):
\[{d\over du}\sin(u)=?\]
OpenStudy (anonymous):
o... from the rule.
cosx=sinx
sinx=-cosx ?
OpenStudy (anonymous):
:D
OpenStudy (anonymous):
those are the rules for integration
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