Q.304. Integration of a trigo function.

Question 304.
Find ∫ sin^2 x cos^4 x dx.

Answer 304.
sin^2 x cos^4 x
= (1/8) * 4sin^2 x cos^2 x * 2cos^2 x
= (1/8) (sin2x)^2 * (1 + cos2x)
= (1/16) (2sin^2 x) (1 + cos2x)
= (1/16) (1 - cos4x) (1 + cos2x)
= (1/16) (1 + cos2x - cos4x - cos4x cos2x)
= (1/32) (2 + 2cos2x - 2cos4x - 2cos4xcos2x)
= (1/32) (2 + 2cos2x - 2cos4x - cos6x - cos2x)
= (1/32) (2 + cos2x - 2cos4x - cos6x)

=> ∫ sin^2 x cos^4 x dx
= (1/32) ∫ (2 + cos2x - 2cos4x - cos6x) dx
= (1/32) (2x + (1/2)sin2x - (1/2)sin4x - (1/6)sin6x) + c
= (1/16) x + (1/64) sin2x - (1/64) sin4x - (1/192) sin6x + c.

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