Question

Find the integrals of the function: $sin^4 x$

Answer 1

sin4 x = sin2 x sin2 x

=($\frac{1-cos2x}{2}$)($\frac{1-cos2x}{2}$)

=$\frac 14$(1-cos2x)2

=$\frac 14$[1+cos22x-2cos2x]

=$\frac 14$[1+($\frac{1-cos4x}{2}$)-2cos2x]

=$\frac 14$[1+$\frac 14$+$\frac 14$cos4x-2cos2x]

=$\frac 14$[$\frac 32$+$\frac 14$cos4x-2cos2x]

∴ ∫sin4 x dx=$\frac 14$ ∫$\frac 32$+$\frac 12$cos4x-2cos2x]dx

=$\frac 14$[$\frac 32$x+$\frac 12$($\frac{sin4x}{4}$)-2sin2x/2]+C

=$\frac 18$[3x+$\frac{sin4x}{4}$-2sin2x]+C

=$\frac{3x}{8}$-$\frac 14$sin2x+$\frac {1}{32}$sin4x+C