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## Homework Statement

Integrate [tex]\int\sin 6x + 3\cos 5x dx[/tex]

## Homework Equations

## The Attempt at a Solution

The way I was taught was the 'u' substitution method.

I know that [tex]\int\sin u du = -\cos u + C[/tex] and I know that [tex]\int\cos u du = \sin u + C[/tex]

Here's what I've done so far

[tex]\int\sin 6x + 3\cos 5x dx[/tex]

[tex]\int\sin 6x dx + 3\int\cos 5x dx[/tex]

u = 6x; u = 5x

du = 6dx; du = 5dx

du/6 = dx; du/5 = dx

[tex]\frac{1}{6}\int\sin u du + \frac{3}{5}\int\cos u du[/tex]

Am I on the right track?

Thanks!

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