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Use trig identities to evaluate indefinite integration of: [4 sin(x) + 4 cos(x)]^2 dx
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16cos(x)-16sin(x)
how'd you find it?
wait what? that's wrong
the question is looking for integration
16(cos(x)+sin(x))^2dx
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16(cos(x)^2+sin(x)^2+2sin(x)cos(x))dx
cos^2+sin^2=1
wait 16(cosx + sinx)^2 is not the same as [4sinx +4cos]^2
no probl. i figured it out but thanks for trying :)
it s same
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because [4(sinx +4cos)]^2 16(sinx +cos)^2
my typing wrong
but answer is 32cos^2-32sin^2 32(1-2sin(x)^2)
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