Asked by INTEGRAL

ʃ (4sin²x cos²×/sin2x cos2x)dx

PLEASE HELP!!

Answers

Answered by Reiny
Your expression can be simplified first.

4sin^2 x cos^2 x /(sin 2x cos 2x)
= 4 sin^2 x cos^2 x/(2sinxcosx cos2x0
= 2sinxcosx/cos2x
= sin 2x/cos 2x

so ʃ (4sin²x cos²×/sin2x cos2x)dx
= ʃ sin2x / cos2x dx
= <b>-(1/2) ln (cos2x) + c </b>

check:

d( -(1/2) ln (cos2x)) /dx
= (-1/2) (1/cos2x)(-2sin2x)
= sin2x/cos2x as needed
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