Integration?

Question

How do you integrate sin^3xcos^3x. I keep during it but getting a weird answer. I wnat to check if I have the right answer or the book does. What substitutions would you use, like sinxcosx= 1/2sin2x ? Please help, midterm is tomorrow.

Answer 1

This is actually one that can be written in several different ways. I am assuming that the sin(x) and the cos(x) are cubed here.

Way 1: rewrite the integrand as [(1/2)sin(2x)]^3=(1/8)sin^3 (2x)=
(1/8)[1-cos^2 (2x)] sin(2x). Now let u=cos(2x), du=-2sin(2x) dx to get
(-1/16)[cos(2x) -(1/3)cos^3 (2x)]+C
=(-1/16)cos(2x)+(1/48)cos^3(2x)+C

Way 2: rewrite as sin^3(x)[1-sin^2(x)]cos(x) and let u=sin(x), du=cos(x)dx. This leads to an anti-derivative of
(1/4)sin^4(x)-(1/6)sin^6(x)+C

Way 3: rewrite as cos^3(x)[1-cos^2(x)] sin(x) and let u=cos(x), du=-sin(x)dx to get an integral of
-(1/4)cos^4(x) +(1/6)cos^6(x)+C

ALL OF THESE ARE CORRECT. They can be interchanged via trig identities and possibly have different constants of integration.

Answer 2

I checked your answer in the next website (very useful):

http://integrals.wolfram.com/index.jsp

if your expression is : sin3x*cos3x the integral is:

-(1/12)* cos 6x

Hope it coincides with the answer in your book.

If this helps you, I can tell you that identity you can use is indeed:

(1/2)sin2x = sinxcosx because in this case:

sin3xcos3x = (1/2)sin 6x

when you integrate this expression you get the answer from the website I search through.


Good luck!

Answer 3

Integration by utilising areas. The function is a manufactured from 2 applications: (x) * (cosx) call considered one of them "u" and the different "dv" opt for the "u" so as that its by-product simplifies u = x du = a million*dx dv = cosx v = sinx that's for the formula setup, that's u*v - crucial (vdu) =xsinx - crucial(sinx * a million * dx) =xsinx - (-cosx) + C = xsinx + cosx + C

Answer 4

let u = sin(3x) then du = 3cos(3x) dx

intergral u*cos(3x)*[du/3cos(3x)] = integral (1/3)* u du

u^2/6 + C = (1/6)* [sin(3x)]^2 + c

Answer 5

Okay, we have:
∫sin³ x cos ³ x dx
∫(sin x cos x)³ dx
∫(sin (2x)/2)³ dx
1/8∫sin³ (2x) dx
1/8∫(sin² (2x)) sin (2x) dx
1/8∫(1- cos² (2x)) sin (2x) dx
1/8∫sin (2x) dx - 1/8∫cos² (2x) sin (2x) dx
-cos (2x)/16 - 1/8 ∫cos² (2x) sin (2x) dx
u=cos (2x), du=-2 sin (2x) dx, sin (2x) dx=-du/2
-cos (2x)/16 + 1/16 ∫u² du
-cos (2x)/16 + u³/48 + C
-cos (2x)/16 + cos³ (2x)/48 + C