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Trig integration HELP PLEASE! step by step:( Ive been trying to solve this one for 2 hours now
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\[\int\limits_{?}^{?}\cos^4xsin^2x\]
indefinite integral
sin and cos are derivatives of each other, so there might be a way to think of this as:\[\int f'(g)g'~dx\to f(g)\]
or using some creative trig identities
but the even exponents might be screwing up the works alittle
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try writing sin^2 x as 1-cos^2 x then you'll have difference of two solvable integrals.
I think youre supposed to use double angle identities but im not sure how.
\[sin(a+a)=sin(a)cos(a)+sin(a)cos(a)\] \[sin(2a)=2sin(a)cos(a)\]
This is how far I got \[\int\limits \frac{ 1 }{ 2 }(1-\cos2x)(\frac{ 1 }{ 2 } \sin2x)^2= \frac{ 1 }{ 8 } \int\limits \sin^2 (2x) -\sin^2 (2x)\cos(2x)\]
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