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

1. anonymous

-3cos(3x)

2. anonymous

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

3. anonymous

oops didnt see the ^

4. anonymous

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

5. anonymous

yes that would be another way of writing the question

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

AH! Forgot something.

17. anonymous

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

18. 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$u=\cos(x), du=-\sin(x)dx, -du=\sin(x)dx$ $-\cos(x)+\int\limits_{}u^2du=-\cos(x)+1/3(\cos^3(x))+c$ $\cos(x)[1/3(\cos^2(x))-1] +c$

19. anonymous

What he said. ^ =D

20. anonymous

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

21. 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?

22. anonymous

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

23. 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.....

24. anonymous

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

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. anonymous

yeah there you go

29. anonymous

Then where did the cos go? from cos^2xsinx I get that u=cos and du = -sinx but what about the other cos?

30. anonymous

bc u=cos(x) $\cos^2(x)= (\cos(x))^2$

31. anonymous

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

32. anonymous

33. 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

34. anonymous

let me know if you are lost