Q:

Evaluate the following integral ∫〖sin〗^4 x〗 dx

0

Evaluate the following integral ∫〖sin〗^4 x〗 dx

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∫〖〖sin〗^4 x〗 dx=∫(〖sin〗^2 x)^2  dx
=∫((1-cos2x)/2)^2  dx
=1/4 ∫(1-2cos2x+〖cos〗^2 2 x)  dx
=1/4 (x-sin2x)+1/4 ∫(1+cos4x)/2 dx
=x/4-1/4 sin2x+1/8 (x+1/4 sin4x)+C
=3/8 x-1/4 sin2x+1/32 sin4x+C

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