How should i integrate (4cos^2 - 1)? Do I use a certain formula...

sin^4cos^2x=(1/16)(1-cos 2x)(1-cos4x) work one 1 side only!

y=4cos(3x-2(3.14))

sin(4x)/sin(x)=4cos(x)cos(x) verify the identity

(cos^2t+4cos^2+4)/(cost+2)=(2sect+1)/sect

5sin(2x)+4cos(x)+0

(3-4cos(2x)+cos(4x))/(3+4cos(2x)+cos(4x))=tan^6(x)

integrate by parts integrate (4+x^2)^1/2

If sin(x) = 4cos(x), then what is (sin(x))*(cos (x))?

4cos^2 x - cosx-3 =0 0 to 2pi SORRY TYPED THE QUESTION WRONG PLESSE HELP

how would i graph g(x)=4cosx

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