MHB Integrating $\sin^3\left({x}\right)\cos\left({x}\right)dx$: Solution

[karush]

$\int\sin^3\left({x}\right)\cos\left({x}\right)dx$
$u=\cos\left({x}\right) du=\sin\left({x}\right)dx$
$\int\left(u-u^3\right)du$
I continued but didn't get the answer

 

[MarkFL]

Since $$\frac{d}{dx}\left(\sin(x)\right)=\cos(x)$$, let:

$$u=\sin(x)\implies du=\cos(x)\,dx$$

 

[karush]

So then...

$\int {u}^{3}du\implies \frac{u^4}{4}\implies\frac{\sin^4\left({x}\right)}{4}+C$

that's too easy!

 

[Prove It]

Too easy? How about this then?

$\displaystyle \begin{align*} \int{\sin^3{(x)}\cos{(x)}\,\mathrm{d}x} &= \int{ \sin^2{(x)}\sin{(x)}\cos{(x)}\,\mathrm{d}x} \\ &= \int{ \frac{1}{2}\left[ 1 - \cos{(2x)} \right] \frac{1}{2}\sin{(2x)}\,\mathrm{d}x} \\ &= \frac{1}{8} \int{ \left[ 1 - \cos{(2x)} \right] \,2\sin{(2x)}\,\mathrm{d}x } \end{align*}$

Now let $\displaystyle \begin{align*} u = 1 - \cos{(2x)} \implies \mathrm{d}u = 2\sin{(2x)}\,\mathrm{d}x \end{align*}$...

What other alternate forms can you come up with?

 

[Krizalid1]

Why doing the second step?

Since $$\dfrac{d}{dx}{{\sin }^{2}}x=2\cos (x)\sin(x)$$

 

[Prove It]

Why doing the second step?

Since $$\dfrac{d}{dx}{{\sin }^{2}}x=2\cos (x)\sin(x)$$

Because I wanted it written in terms of (2x) :P