Question

integrate sin^4x cos^4x dx
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Answer 1

Answer:

sin4xcos4x=116(16sin4xcos4x)=116(2sinxcosx)4sin4xcos4x=116(16sin4xcos4x)=116(2sinxcosx)4

=116(sin2x)4=116(sin22x)2=116(1−cos22x)2∵2sinαcosα=sin2αandsin2θ=1−cos2θ=116(sin2x)4=116(sin22x)2=116(1−cos22x)2∵2sin⁡αcos⁡α=sin2αandsin2θ=1−cos2θ

=116(1+cos42x−2cos22x)=116(1+cos42x−2cos22x)

=116+116cos42x−18cos22x=116+116cos42x−18cos22x

=116+116(cos22x)2−18(1+cos4x2)∵cos2θ=1+cos2θ2=116+116(cos22x)2−18(1+cos4x2)∵cos2θ=1+cos2θ2

=116+116(1+cos4x2)2−116−116cos4x=116+116(1+cos4x2)2−116−116cos4x

=164(1+cos24x+2cos4x)−116cos4x=164(1+cos24x+2cos4x)−116cos4x

=164+164cos24x+132cos4x−116cos4x=164+164cos24x+132cos4x−116cos4x

=164+164cos24x−132cos4x=164+164cos24x−132cos4x

=164+164(1+cos8x2)−132cos4x=164+164(1+cos8x2)−132cos4x

=164+1128+1128cos8x−132cos4x=164+1128+1128cos8x−132cos4x

=3128+1128cos8x−132cos4x=3128+1128cos8x−132cos4x

∫sin4xcos4xdx∫sin4xcos4xdx

=∫(3128+1128cos8x−132cos4x)dx=∫(3128+1128cos8x−132cos4x)dx

=3128x+11024sin8x−1128sin4x+c=3128x+11024sin8x−1128sin4x+c

=11024(24x+sin8x−8sin4x)+c=11024(24x+sin8x−8sin4x)+c

=11024(24x−8sin4x+sin8x)+c=11024(24x−8sin4x+sin8x)+c

Answer 2

Answer:

this ans is correct