Question
ʃ (4sin²x cos²×/sin2x cos2x)dx
PLEASE HELP!!
PLEASE HELP!!
Answers
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
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