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Question

What is the integral of sin4(x)?


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Solution

Find the integral of sin4(x)

Reframe and further calculate the integral:

sin4xdx=sin2x2dxsin4xdx=1-cos2x22dx{sin2x=1-cos2x2}sin4xdx=141+cos22x-2cos2xdxsin4xdx=141+1+cos4x2-2cos2xdx{cos22x=1+cos4x2}sin4xdx=143x2+sin4x8-2sin2x2+C{cosθ=sinθ.Also,Cisaconstant}sin4xdx=3x8+sin4x32-sin2x4+C

Hence, sin4xdx=3x8+sin4x32-sin2x4+C


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