Math, asked by smithdestiny3731, 1 year ago

∫sin2xdxa2cos2x+b2sin2xsin2x=(a)(b−a)log(a2cos2x+b2sin2x)(b)1b−alog(a2cos2x+b2sin2x)(c)1b2−a2log(a2cos2x+b2sin2x)(d)1a2+b2log(a2cos2x+b2sin2x)

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Answered by Anonymous

0

heya...

Option D is the answer

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