how do you find the integral of sin^3x?

- anonymous

how do you find the integral of sin^3x?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

-3cos(3x)

- anonymous

the answer is -cosx + (cos^3x/3) + c
how do you get there?

- anonymous

oops didnt see the ^

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

so then im guessing what its SUPPOSED to be is sin(x)^3?

- anonymous

yes that would be another way of writing the question

- anonymous

\[\int\limits_{}^{} \sin^3xdx=\int\limits_{}^{}(1-\cos^2x)(sinx)dx\]

- anonymous

So then use then let u=cosx, so then du=-sinxdx.

- anonymous

What I do not understand is why u=cosx?! How did you know to choose that?

- anonymous

So then you have, \[-\int\limits_{}^{}1-u^2du\]

- anonymous

Essentially you want to get it to a point where you can work with it.

- anonymous

Since I want to get rid of the sin stuff, so to speak, I substitute cosx to get sinx grouped with dx

- anonymous

It's harder to explain via computer though. :/

- anonymous

I'm assuming you can take the integral of the one I typed, correct?

- anonymous

yes, but where did the 1 from 1- u^2 come from? Sorry, this is hard for me to grasp

- anonymous

You could have also done it this way: \[\int\limits_{}^{} \sin^3xdx=\int\limits_{}^{}(1-\sin^2x)(cosx)dx\]

- anonymous

AH! Forgot something.

- anonymous

You could have substituted 1-cosx^2 for u as well

- 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

What he said. ^ =D

- anonymous

sorry mistake
\[\int\limits_{}\sin(x)dx=\int\limits_{}\cos^2(x)\sin(x)dx\]
then u substitution

- 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

no that is the final answered I factored out a cos(x) because both terms in the final answer have cos(x)

- 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

if u = cosx then where did the cosxsinx go? I am so confusedd

- anonymous

because - -cos3x/3 = +cos3x/3 two negatives in math make a positive

- anonymous

sin(x) was taken care of by substituting -du=sinxdx

- anonymous

ohh is that where the (-1) came from? from -du = sinx?

- anonymous

yeah there you go

- 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

bc u=cos(x)
\[\cos^2(x)= (\cos(x))^2\]

- anonymous

Ohhh thank you sooo much to everyone who helped!! I get it now ! :D

- anonymous

your welcome...... hope this helped

- 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

let me know if you are lost

Looking for something else?

Not the answer you are looking for? Search for more explanations.