How do you find the differential of the integral cos^3(x) dx from -9ts to -5t+6s?

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How do you find the differential of the integral cos^3(x) dx from -9ts to -5t+6s?

Mathematics
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\[\int\limits_{-9ts}^{-5t+6s}\cos ^{3}(x) dx\]
ummmm......... i dont really know...
i spose its possible, just use those"equations" as the F(x) stuff in the end

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Other answers:

you might want to go about "reducing" the integrand to an addition of sin and cos....maybe
cos cos^2 cos(1-sin^2) cos(x)(1 - (1-cos(2x)/2)) ... along those lines eh?
yeah thats what i was going to say
well say it again then :)
integral of sin^2 is known as -x/2 + sincos/2 so you could use that
Well I have notes and an example problem, but it doesnt really make sense. He has the differential of the integral e^-x^2 from a to b. He then takes partials. Fa = -e^a^2 and Fb = e^b^2. Then apparently the differential is -e^a^2 * da + e^b^2 * db. No sense.
Don't worry about it, I just wondered if anyone could make sense out of it, but obviously my teacher is drunk. He wrote his own "textbook" in Mathematica and he didn't do a very good job.
are you still talking about integral of cos^3? anyway yeah there are no elementary functions to describe integral of e^(-x^2).

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