Use the product to sum identity:
Then you'll get a simpler integral.
Ok so here is my integral:
INTEGRAL of [sin(5x)][cos(6x)]dx
Now this is a pretty simple u-substitution problem, i know that. Except i have one problem. The 5x and 6x inside the two trig functions is throwing me off. Is there a way to pull them apart so i just have a whole bunch of sin(x)'s and cos(x)'s? I'm stumped.
No problem.
Visual Calculus - Trigonometric Identities
For my answer I got:
-.5[ (1/11)cos(11x) + cos(-x)]
Can anyone confirm this for me?
My work is as follows:
INTEGRAL of: Sin5x * Cos6x = 1/2 * INTEGRAL of: sin(11x)+sin(-x)
by virtue of the Product sum identity
Which simplifies to -.5[ (1/11)cos(11x) + cos(-x)]
Hello,
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The Ò is for the integral sign.
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