- #1

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## Homework Statement

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

[itex]\int sin^{3}7x dx[/itex]

## Homework Equations

## The Attempt at a Solution

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

Here's where the trouble begins.

[itex]\frac{1}{7}\int sin^{3}U dU[/itex]

Now my book claims the following:

[itex]\int sin^{n}x dx = -\frac{1}{n}sin^{n-1}x cos x + \frac{n-1}{n}\int sin^{n-2}xdx[/itex]

Working that gives me:

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

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

[itex]\int sin^{n}x dx = -\frac{1}{n}sin^{n-1}x cos x + \frac{n-1}{n}\int sin^{n-2}xdx[/itex]

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