Integrate Trig Functions with Ease

[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

Help, please, and be blessed!

(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?)

 

[courtrigrad]

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

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

 

[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

 

[courtrigrad]

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?

 

[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
$$\int(u^3-u^5)(-1)du$$
$$=\int(u^5-u^3)du$$
$$=\frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x} + C$$

 

[courtrigrad]

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

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

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

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

 

[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?

 

[courtrigrad]

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

 

[mbrmbrg]

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

 

[courtrigrad]

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

 

[HallsofIvy]

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

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 $\pi/4$.

 

[courtrigrad]

the integral between the limits a and b would be the same, even though the actual functions are not equal to each other. that's 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 $$\frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x}$$

 

[mbrmbrg]

the integral between the limits a and b would be the same, even though the actual functions are not equal to each other. that's the key idea.

That is so cool!
But why, please?

And thank you very much for your prolonged patience.

 

[courtrigrad]

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.

 

[HallsofIvy]

That is so cool!
But why, please?

And thank you very much for your prolonged patience.

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

 

[mbrmbrg]

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.

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!