1. ## A simple integral

Hi,

I'd be happy if someone could check my work on the integral $sin^6(x)dx$

$sin^6(x) = (sin^2(x))^3$

$sin^2(x) = \frac{1}{2}(1-cos(2x))$

$\frac{1}{8}\int(1-cos(2x))^3$

let u = cos(2x);

$(1-u)^2\cdot(1-u) = 1-3u+3u^2-u^3$

$= \frac{1}{8}\int~1-3cos(2x)+3cos^2(2x)-cos^3(2x)$

= $3~\int cos^2(2x) = \frac{1}{2}(1+cos(2x)) = \frac{1}{2}x+\frac{1}{4}sin(2x)$

2. first when u use substitution let's say "u" in your case, where is your "du" ?

3. Dang, you're right, how do i deal with that?

4. Originally Posted by Jones
Dang, you're right, how do i deal with that?
u put $u = \cos{2x}$ so "du" is derivation of "u" so $du = -2 \sin {2x}$

5. You also can prove the reduction formula, $\int sin^m(x) dx = \frac{-(cos(x) sin^{(m-1)}(x))}{m} + \frac{(m-1)}{m} \int sin^{(-2+m)}(x) dx$

And use it 3 times when your $m$ is $6$.

6. Originally Posted by yeKciM
u put $u = \cos{2x}$ so "du" is derivation of "u" so $du = -2 \sin {2x}$
Jones can't do that because in what he/she had:
$\frac{1}{8}\int(1-cos(2x))^3$
there is no sin 2x.

Jones: from here:
$\frac{1}{8}{\displaystyle{\int}}(1-cos(2x))^3 dx$
multiply out (1-cos2x)^3 to get

$\frac{1}{8}{\displaystyle{\int}} (1 - 3 \cos 2x + 3\cos^2 2x - \cos^3 2x)\, dx$

Separate into four integrals. The first two are straight forward. For the third, re-apply the power-reduction identity
$\cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}$.

For the fourth, note that
$\cos^3 \theta = \cos^2 \theta \cdot \cos \theta = (1 - \sin^2 \theta)\cos \theta$.
Use this to evaluate the 4th integral.