Integrating sin³(7x) using reduction formula

[1MileCrash]

Homework Statement

This is perhaps the most confused I've ever been.


$\int sin^{3}7x dx$

Homework Equations

The Attempt at a Solution

First, I let u = 7x. du = 7 dx

Here's where the trouble begins.


$\frac{1}{7}\int sin^{3}U dU$

Now my book claims the following:

$\int sin^{n}x dx = -\frac{1}{n}sin^{n-1}x cos x + \frac{n-1}{n}\int sin^{n-2}xdx$

Working that gives me:

$\frac{1}{7}[-\frac{1}{3}sin^{2}u cos u - \frac{2}{3} cos u]$

Here's the problem. Not only does this simplify to a completely incorrect answer, but my application of:

$\int sin^{n}x dx = -\frac{1}{n}sin^{n-1}x cos x + \frac{n-1}{n}\int sin^{n-2}xdx$

Doesn't even lead to a correct integral of (sin x)^3 alone, according to wolfram. Is this formula BS?

 

[LCKurtz]

Why don't you start by writing

$$\sin^3(7x) = \sin(7x)(1 - cos^2(7x)$$

You don't need a reduction formula at all.

 

[1MileCrash]

That's all well and good, but I really need to understand why what I did does not work.

 

[ehild]

Integrating by parts mean that

$$\int{u'vdx}=uv-\int{uv' dx}$$

$$\int{sin^n(x)}dx =\int{sin(x)sin^{n-1}(x)dx}$$,

u'=sin(x) and v=sin^(n-1)(x).

$$\int{sin(x)sin^{n-1}(x)dx}=-cos(x)sin^{n-1}(x)+(n-1)\int{cos^2(x)sin^{n-2}(x)dx}$$

Edit: Replacing cos^2(x) by 1-sin^2(x) and rearranging, it leads to your formula.

$$\int sin^{n}x dx = -\frac{1}{n}sin^{n-1}x cos x + \frac{n-1}{n}\int sin^{n-2}xdx$$

ehild

 

[ArcanaNoir]

Your formula is correct. My book also lists the formula
$$\int{sin^3udu}=-\frac{1}{3}(2+sin^2u)cosu+C$$
Which is what I got when I used the generic $n$ formula.

Perhaps you simplified incorrectly. (your work posted here appears correct) And perhaps what wolfram returned was equivalent to the known answer.