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Question Number 54971 by peter frank last updated on 15/Feb/19
∫0π2tanxsin2xsinxcosxtanxdx
Answered by kaivan.ahmadi last updated on 15/Feb/19
∫0π2sinxcosxsin2xsinxcosxsinxcosxdx=∫0π2sinxdx=−cosx]0π2=1
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Feb/19
∫tanx×tan2x×cos2xtanx×cos2x×tanxdx∫tanx×tanx×cosxtanxdx∫sinxdx=−∣cosx∣0π2=−(0−1)=1ans
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