MHB Where Does the $\frac{1}{2}$ in Step 3 Come From?

[karush]

https://www.physicsforums.com/attachments/5246
I'm trying to follow this example but where does the $\frac{1} {2 } $ come from in step 3

 

[MarkFL]

They are letting:

$$u=\sin(2x)\implies \frac{1}{2}du=\cos(2x)\,dx$$

Does that make sense?

 

[karush]

Sorta, So to get the 2 they multiplied by $\frac{1}{2}$
Letters me try one

$$\int\sin^3\left({x}\right) dx $$

$$\int\left(\sin^2\left({x}\right)\right)\left(\sin\left({x}\right)\right)dx
\implies \int\left(\cos^2\left({x}\right)-1\right) \left(\sin\left({x}\right)\right)dx $$

$$u=\sin\left({x}\right)\ \ \ du =\cos\left({x}\right)dx$$

So far?

 

[Prove It]

Sorta, So to get the 2 they multiplied by $\frac{1}{2}$
Letters me try one

$$\int\sin^3\left({x}\right) dx $$

$$\int\left(\sin^2\left({x}\right)\right)\left(\sin\left({x}\right)\right)dx
\implies \int\left(\cos^2\left({x}\right)-1\right) \left(\sin\left({x}\right)\right)dx $$

$$u=\sin\left({x}\right)\ \ \ du =\cos\left({x}\right)dx$$

So far?

You substituted the wrong function. You want to substitute whatever is INSIDE the composition, not what is outside, so you would want u = cos(x) and du = -sin(x) dx.

Also [sin(x)]^2 = 1 - [cos(x)]^2, not [cos(x)]^2 - 1...

 

[karush]

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

$$u=\cos\left({x}\right)\ \ \ du=-\sin\left({x}\right)dx $$

$$-\int\left(1-{u}^{2}\right)du $$

I continued but didn't get this TI-Nspire answer

$$\left(\frac{-\sin^2 \left({x}\right)}{3}-\frac{2}{3}\right)\cos\left({x}\right)$$

 

[Prove It]

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

$$u=\cos\left({x}\right)\ \ \ du=-\sin\left({x}\right)dx $$

$$-\int\left(1-{u}^{2}\right)du $$

I continued but didn't get this TI-Nspire answer

$$\left(\frac{-\sin^2 \left({x}\right)}{3}-\frac{2}{3}\right)\cos\left({x}\right)$$

What you have done is correct, so keep going.

 

[karush]

$$- \int 1 du + \int {u^{2 }} du $$

$$-u+\frac{{u}^{3}}{3}+C$$

 

[Prove It]

So now put it back as a function of x. You should be able to see why the two answers are equivalent.

 

[karush]

$$-\cos\left({x}\right)-\frac{\cos^3\left({x}\right)}{3}
\implies\left[-1-\frac{\cos^2 \left({x}\right)}{3}\right]\cos\left({x}\right)
\implies\left[\frac{-3+1+\sin^2 \left({x}\right)}{3}\right]\cos\left({x}\right)$$

Thus

$$\left[\frac{-\sin^2 \left({x}\right)}{3}-\frac{2}{3}\right]\cos\left({x}\right)$$

 

[suluclac]

$$-\int\cos^3{2x}\,dx=-\frac{1}{4}\int\cos{6x}\,dx-\frac{3}{4}\int\cos{2x}\,dx=-\frac{1}{4}(\frac{1}{6})\sin{6x}-\frac{3}{4}(\frac{1}{2})\sin{2x}+C$$
$$\therefore-\int\cos^3{2x}\,dx=-\frac{1}{24}\sin{6x}-\frac{3}{8}\sin{2x}+C$$
This, of course, requires the reader the ability to prove $$\cos{6x}=4\cos^3{2x}-3\cos{2x}$$.