- #1
1MileCrash
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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?