# Calculus

integrate (pi&0) sin t/(2-cost) dt

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1. My calculation:
u=2-cos t
du/dt=sin t
du=sin t dt
When t=pi,
u=2-cos pi=1.002
t=0, u=2-cos 0=1
integral (1.002&1) (1/u) du
=(ln|u|) (1.002&1)
stucked

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2. I noticed the following ...

A denominator, whose derivative shows up in the numberator...
Ahhh, logs!

∫ sin t/(2-cost) dt
= ln (2-cost)

take it from there.

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posted by Reiny
3. and remember that log 3 is just a number, like any other. Don't worry about evaluating it as a decimal. Who cares what the decimal approximation is? Just like √7 or 3/π, it's fine to leave it as it is.

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posted by Steve

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