1. ## Another Trig Antiderivative

Here's another one for you folks:

$\int sin^3(x)cos^3(x)dx$

I have no idea which trigonometric identity to use... could you guys give me an idea.

I feel like the only one might possibly be [tex] sin(x)cos(y) = 1/2(sin(x+y) + sin(x-y)) but I don't feel like this is the right one.

Please tell me I'm not just over thinking it... could you do this? I'm also 100% sure you couldn't do this but just making sure.
$\int (sin(x)cos(x))^3$ equals $\frac {(sin(x)cos(x))^4}{4}$?

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

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

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

Now, let $u=sin(x), \;\ du=cos(x)dx$ and it falls right into place.

3. Originally Posted by galactus
$\int{sin^{3}(x)cos^{3}(x)}dx$

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

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

Now, let $u=sin(x), \;\ du=cos(x)dx$ and it falls right into place.
Ok so is this all right...

$\int (u^3 - u^5) cos(x)dx$

$(\frac {u^4}{4} - \frac {u^6}{6}) sin(x)$

then just replace u with sin(x)...

$(\frac {sin^4(x)}{4} - \frac {sin^6(x)}{6}) sin(x)$

could you then:
$(\frac {sin^5(x)}{4} - \frac {sin^7(x)}{6})$

4. Originally Posted by larson
Ok so is this all right...

$\int (u^3 - u^5) cos(x)dx$

$(\frac {u^4}{4} - \frac {u^6}{6}) sin(x)$

then just replace u with sin(x)...

$(\frac {sin^4(x)}{4} - \frac {sin^6(x)}{6}) sin(x)$

could you then:
$(\frac {sin^5(x)}{4} - \frac {sin^7(x)}{6})$
Close, but not quite. You can't take the integral of $f(u)$ with respect to $x$. So, since we substituted $u$ for $sin(x)$, you also need to get rid of $cos(x)$ and $dx$ for something in terms of $u$.

We know:

$u=sin(x)$

So, take the derivative of that to get:

$du/dx=cos(x)$

$du=cos(x)dx$

5. Originally Posted by hatsoff
Close, but not quite. You can't take the integral of $f(u)$ with respect to $x$. So, since we substituted $u$ for $sin(x)$, you also need to get rid of $cos(x)$ and $dx$ for something in terms of $u$.

We know:

$u=sin(x)$

So, take the derivative of that to get:

$du/dx=cos(x)$

$du=cos(x)dx$

$(\frac {sin^4(x)}{4} - \frac {sin^6(x)}{6}) \frac {sin^3(x)}{3}$?

which can then equal:
$(\frac {sin^7(x)}{12} - \frac {sin^9(x)}{18})$???

6. No, I am sorry to say. It's right in front of you. As stated before, let

$u=sin(x). \;\ du=cos(x)dx$

$\int{(u^{3}-u^{5})}du$

The du takes care of the cos(x)dx.

Now integrate and resub

$\frac{1}{4}u^{4}-\frac{1}{6}u^{6}$

Now, resub:

$\frac{1}{4}sin^{4}(x)-\frac{1}{6}sin^{6}(x)+C$

7. Originally Posted by larson

$(\frac {sin^4(x)}{4} - \frac {sin^6(x)}{6}) \frac {sin^3(x)}{3}$?

which can then equal:
$(\frac {sin^7(x)}{12} - \frac {sin^9(x)}{18})$???
You're making the same mistake you did before. Recall your first substitution:

$\int (u^3 - u^5) cos(x)dx$

That was correct! But you still need to do more; you need to either *completely* evaluate the integral in terms of $x$ (and you can't do that), or *completely* evaluate the integral in terms of $u$.

In other words, you need to get rid of all $x$ and $dx$ terms.

8. Originally Posted by hatsoff
You're making the same mistake you did before. Recall your first substitution:

$\int (u^3 - u^5) cos(x)dx$

That was correct! But you still need to do more; you need to either *completely* evaluate the integral in terms of $x$ (and you can't do that), or *completely* evaluate the integral in terms of $u$.

In other words, you need to get rid of all $x$ and $dx$ terms.
Ok so, tell me if I'm doing this right then...
$\int sin^3(x)cos^3(x)dx$
equals:
$\int sin^3(x)(1 - sin^2(x))cosx dx$
let $u = sin(x)$ so $du = cos(x)dx$ so $dx = \frac {du}{cos(x)}$

this all equals:
$\int (u^3 - u^5)cos(x) \frac {du}{cos(x)}$
both of the cos(x)'s cancel each other out so your left with:

$\int (u^3 - u^5)$

equals:

$\frac{u^4}{4} - \frac {u^6}{6}$
then replace the u's with sin(x)...

$\frac{sin^4(x)}{4} - \frac {sin^6(x)}{6}$

and if you wanted to make it look a little nicer...
$\frac{3sin^4(x) - 2sin^6(x)}{12}$
I hope this is right or else I'm done with haha.

9. Originally Posted by larson
Ok so, tell me if I'm doing this right then...
$\int sin^3(x)cos^3(x)dx$
equals:
$\int sin^3(x)(1 - sin^2(x))cosx dx$
let $u = sin(x)$ so $du = cos(x)dx$ so $dx = \frac {du}{cos(x)}$

this all equals:
$\int (u^3 - u^5)cos(x) \frac {du}{cos(x)}$
both of the cos(x)'s cancel each other out so your left with:

$\int (u^3 - u^5)$

equals:

$\frac{u^4}{4} - \frac {u^6}{6}$
then replace the u's with sin(x)...

$\frac{sin^4(x)}{4} - \frac {sin^6(x)}{6}$
Correct!

and if you wanted to make it look a little nicer...
$\frac{3sin^4(x) - 2sin^6(x)}{12}$
I hope this is right or else I'm done with haha.
Actually, once you get to calculus, algebraic simplification isn't required. That is, it looked nicer before. But you're still correct.

10. Originally Posted by hatsoff
Correct!

Actually, once you get to calculus, algebraic simplification isn't required. That is, it looked nicer before. But you're still correct.
Well guys, thanks so much for helping me out.