anonymous
  • anonymous
how do you find the integral of sin^3x?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
-3cos(3x)
anonymous
  • anonymous
the answer is -cosx + (cos^3x/3) + c how do you get there?
anonymous
  • anonymous
oops didnt see the ^

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anonymous
  • anonymous
so then im guessing what its SUPPOSED to be is sin(x)^3?
anonymous
  • anonymous
yes that would be another way of writing the question
anonymous
  • anonymous
\[\int\limits_{}^{} \sin^3xdx=\int\limits_{}^{}(1-\cos^2x)(sinx)dx\]
anonymous
  • anonymous
So then use then let u=cosx, so then du=-sinxdx.
anonymous
  • anonymous
What I do not understand is why u=cosx?! How did you know to choose that?
anonymous
  • anonymous
So then you have, \[-\int\limits_{}^{}1-u^2du\]
anonymous
  • anonymous
Essentially you want to get it to a point where you can work with it.
anonymous
  • anonymous
Since I want to get rid of the sin stuff, so to speak, I substitute cosx to get sinx grouped with dx
anonymous
  • anonymous
It's harder to explain via computer though. :/
anonymous
  • anonymous
I'm assuming you can take the integral of the one I typed, correct?
anonymous
  • anonymous
yes, but where did the 1 from 1- u^2 come from? Sorry, this is hard for me to grasp
anonymous
  • anonymous
You could have also done it this way: \[\int\limits_{}^{} \sin^3xdx=\int\limits_{}^{}(1-\sin^2x)(cosx)dx\]
anonymous
  • anonymous
AH! Forgot something.
anonymous
  • anonymous
You could have substituted 1-cosx^2 for u as well
anonymous
  • anonymous
\[\int\limits_{}\sin^3(x)dx=\int\limits_{}\sin^2(x)\sin(x)dx\] \[\int\limits_{}(1-\cos^2(x))\sin(x)dx\] \[\int\limits_{}\sin(x)dx-\int\limits{}\cos^2(x)dx\] use u substitution\[u=\cos(x), du=-\sin(x)dx, -du=\sin(x)dx\] \[-\cos(x)+\int\limits_{}u^2du=-\cos(x)+1/3(\cos^3(x))+c\] \[\cos(x)[1/3(\cos^2(x))-1] +c\]
anonymous
  • anonymous
What he said. ^ =D
anonymous
  • anonymous
sorry mistake \[\int\limits_{}\sin(x)dx=\int\limits_{}\cos^2(x)\sin(x)dx\] then u substitution
anonymous
  • anonymous
cos(x)[1/3(cos2(x))−1]+c I don't understand this step Where does the -1 come from? this is before the final answer right?
anonymous
  • anonymous
no that is the final answered I factored out a cos(x) because both terms in the final answer have cos(x)
anonymous
  • anonymous
if du = -sinxdx and u= cosx Then you make it -cosx - integral of u^2 thennn you get -cosx -cos3x/3? how did you get -1 to make it -cosx+cos3x.....
anonymous
  • anonymous
if u = cosx then where did the cosxsinx go? I am so confusedd
anonymous
  • anonymous
because - -cos3x/3 = +cos3x/3 two negatives in math make a positive
anonymous
  • anonymous
sin(x) was taken care of by substituting -du=sinxdx
anonymous
  • anonymous
ohh is that where the (-1) came from? from -du = sinx?
anonymous
  • anonymous
yeah there you go
anonymous
  • anonymous
Then where did the cos go? from cos^2xsinx I get that u=cos and du = -sinx but what about the other cos?
anonymous
  • anonymous
bc u=cos(x) \[\cos^2(x)= (\cos(x))^2\]
anonymous
  • anonymous
Ohhh thank you sooo much to everyone who helped!! I get it now ! :D
anonymous
  • anonymous
your welcome...... hope this helped
anonymous
  • anonymous
I made a typing error again.... here is the whole thing without mistakes \[∫\sin^3(x)dx=∫\sin^2(x)\sin(x)dx\] \[∫(1−\cos^2(x))\sin(x)dx\] \[∫\sin(x)dx-∫\cos^2(x)\sin(x)dx\] \[u=\cos(x),du=−\sin(x)dx,−du=\sin(x)dx\] \[−\cos(x)+∫u^2du=−\cos(x)+1/3(\cos^3(x))+c\] \[\cos(x)[1/3(\cos^2(x))−1]+c\] sorry for the confusion
anonymous
  • anonymous
let me know if you are lost

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