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Question Number 88045 by jagoll last updated on 08/Apr/20
∫sinx2sinx+cosxdx
Commented by john santu last updated on 08/Apr/20
sinx2sinx+cosx=25−15[2cosx−sinx2sinx+cosx]⇒∫{25−15[2cosx−sinx2sinx+cosx]}dx=2x5−15log∣2sinx+cosx∣+c
Commented by jagoll last updated on 08/Apr/20
waw...supereasy
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