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anonymous
 5 years ago
how do you find the integral of sin^3x?
anonymous
 5 years ago
how do you find the integral of sin^3x?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is cosx + (cos^3x/3) + c how do you get there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then im guessing what its SUPPOSED to be is sin(x)^3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes that would be another way of writing the question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{} \sin^3xdx=\int\limits_{}^{}(1\cos^2x)(sinx)dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So then use then let u=cosx, so then du=sinxdx.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What I do not understand is why u=cosx?! How did you know to choose that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So then you have, \[\int\limits_{}^{}1u^2du\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Essentially you want to get it to a point where you can work with it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since I want to get rid of the sin stuff, so to speak, I substitute cosx to get sinx grouped with dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's harder to explain via computer though. :/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm assuming you can take the integral of the one I typed, correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but where did the 1 from 1 u^2 come from? Sorry, this is hard for me to grasp

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You could have also done it this way: \[\int\limits_{}^{} \sin^3xdx=\int\limits_{}^{}(1\sin^2x)(cosx)dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0AH! Forgot something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You could have substituted 1cosx^2 for u as well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}\sin^3(x)dx=\int\limits_{}\sin^2(x)\sin(x)dx\] \[\int\limits_{}(1\cos^2(x))\sin(x)dx\] \[\int\limits_{}\sin(x)dx\int\limits{}\cos^2(x)dx\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0sorry mistake \[\int\limits_{}\sin(x)dx=\int\limits_{}\cos^2(x)\sin(x)dx\] then u substitution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos(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
 5 years ago
Best ResponseYou've already chosen the best response.0no that is the final answered I factored out a cos(x) because both terms in the final answer have cos(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0if u = cosx then where did the cosxsinx go? I am so confusedd

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because  cos3x/3 = +cos3x/3 two negatives in math make a positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sin(x) was taken care of by substituting du=sinxdx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh is that where the (1) came from? from du = sinx?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then where did the cos go? from cos^2xsinx I get that u=cos and du = sinx but what about the other cos?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bc u=cos(x) \[\cos^2(x)= (\cos(x))^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ohhh thank you sooo much to everyone who helped!! I get it now ! :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your welcome...... hope this helped

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0let me know if you are lost
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