Homework Help: Integration with trig

1. Oct 12, 2006

mbrmbrg

(1)$$\int\sin^3{x}\cos^3{x}dx$$
(2)$$=\int\sin^3{x}(1-\sin^2{x})\cos{x}dx$$
(3a) let u=sin(x), du=cos(x)dx
(3b)$$=\int(u^3-u^5)du$$
(4)$$=\frac{u^4}{4} -\frac{u^6}{6} + C$$
(5)$$=\frac{1}{4}\sin^4{x} -\frac{1}{6}\sin^6{x} + C$$

If at step (2) I pull off a sin(x) and go from there, I end up with $$\frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x} + C$$
I tried to prove that this is equvalent to step (5) by using sin^2(A)+cos^2(A)=1, but this turned step (5) into $$\frac{1}{12} +\frac{3}{4}\cos^4{x} -\frac{1}{6}\cos^6{x} + C$$. And so I was bewildered and befuddled.

Then I decided to plug the original problem into a website's Integrator. And the integrator told me that the answer really is $$\frac{1}{192}(-9\cos(2x) +\cos(6x))$$.

Now I am just

(Last little bit of frustration: I typed up this post before I logged in, and even the browser's back button couldn't help me retrieve it. Latex is cool but twice?)

2. Oct 12, 2006

So your doing $$\int \sin x(\sin^{2} x - \sin^{4} x) dx$$?

That is equaled to $$\int(u^{3}-u^{5}) du$$

3. Oct 12, 2006

mbrmbrg

yes, because what I did was
$$\int\sin^3{x}(1-\sin^2{x})(\cos{x})dx$$
$$=\int(\sin^3{x}-\sin^5{x})(\cosx)dx$$
$$=\int(u^3-u^5)du$$
if u=sinx, du=cosxdx

Last edited: Oct 12, 2006
4. Oct 12, 2006

ok. so what are you having problems with then? $$\int \sin x(\sin^{2} x - \sin^{4} x) dx$$ is equivalent do what you did. you say you pulled off a $$\sin x$$. Did you mean factor it like I did in the integral?

5. Oct 12, 2006

mbrmbrg

No, I mean that instead of my original step (2), I did this:
$$\int(1-\cos^2{x})(\cos^3{x})(\sin{x})dx$$
$$=\int(\cos^3{x}-\cos^5{x})(-1)(-\sin{x})$$
let u=cos(x), du=-sin(x), so the integral becomes
$$\int(u^3-u^5)(-1)du$$
$$=\int(u^5-u^3)du$$
$$=\frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x} + C$$

6. Oct 12, 2006

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

$$\int(\cos^{3} x - \cos^{5}x) \sin x dx$$

$$u = \cos x, du = -\sin x$$.

$$-\int u^{3} - u^{5} du = -(\frac{\cos^{4}x}{4}-\frac{\cos^{6}x}{6})$$
$$\int u^{3} - u^{5}du = (\frac{\cos^{4}x}{4}-\frac{\cos^{6}x}{6})$$

7. Oct 12, 2006

mbrmbrg

That's exactly it--
If I set u=sin(x) I seem to get a different answer that if I set u=cos(x).
But why?

8. Oct 12, 2006

They are equivalent because of the constant of integration. The constants are $$C_{1}, C_{2}$$

9. Oct 16, 2006

mbrmbrg

But if I wanted to evaluate the integral between a and b, wouldn't the constants cancel?

10. Oct 16, 2006

The integral should be $$\frac{\cos^{6}x}{6}-\frac{\cos^{4}x}{4}$$

11. Oct 16, 2006

HallsofIvy

Yes, they would. Did you try it?
For example, you could have
$$\int_0^{\frac{pi}{2}}sin^4x cos^3xdx= \frac{1}{4}sin^4x- \frac{1}{6}sin^6x+ C\left|_0^{\frac{\pi}{2}}$$
or
$$\int_0^{\frac{pi}{2}}sin^4x cos^3xdx= \frac{1}{6}cos^6x-\frac{1}{4}cos^4x+ C\left|_0^{\frac{\pi}{2}}$$
Those are exactly the same. For a slightly harder example try between 0 and $\pi/4$.

12. Oct 16, 2006

the integral between the limits a and b would be the same, even though the actual functions are not equal to each other. thats the key idea. For example:

$$\int_{0}^{1} \frac{1}{4}\sin^4{x} -\frac{1}{6}\sin^6{x} = \int_{0}^{1} \frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x}$$ even though $$\frac{1}{4}\sin^4{x} -\frac{1}{6}\sin^6{x}$$ does not equal $$\frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x}$$

Last edited: Oct 16, 2006
13. Oct 16, 2006

mbrmbrg

That is so cool!!

And thank you very much for your prolonged patience.

14. Oct 16, 2006

you could look at the graph, and see that the areas are the same (i.e. area below x-axis and area above x-axis). maybe employ some trig identities. Also look at the graphs of sin(x) and cos(x) together.

Last edited: Oct 16, 2006
15. Oct 16, 2006

HallsofIvy

Pick specific values for a and b, do the calculation yourself and see why!

16. Oct 16, 2006

mbrmbrg

Ohhhhhh.... :understanding dawns:
I love math.:!!)
Hail PhysicsForums, bastion of nerdiness!

And yes, HallsofIvy, I did evaluate the integral between 0 and pi/2 (took the easy way out because I just wanted to see what was going on), and I saw the coolness take place before my very eyes, but I still understood it not. Sigh.

Pictures are great. Graphs are pictures. ==> Graphs are great.

Thank you, people!