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Question Number 21707 by Isse last updated on 01/Oct/17
∫π/6π/312cot22θdθ
Commented by Tikufly last updated on 01/Oct/17
I=12∫π/6π/3(cosec22θ−1)dθ=12∫π/6π/3cosec22θ−12∫π/6π/31dθ=−14[cot2θ]π/6π/3−12[θ]π/6π/3=−14{cot(2θ3)−cot(θ3)}−12(π3−π6)=−14(−13−13)−12(π6)=123−π12
Commented by Isse last updated on 01/Oct/17
thanks
Commented by Tikufly last updated on 02/Oct/17
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