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Question Number 21707 by Isse last updated on 01/Oct/17

∫_(π/6) ^(π/3) (1/2)cot^2 2θdθ

π/6π/312cot22θdθ

Commented by Tikufly last updated on 01/Oct/17

I=(1/2)∫_(π/6) ^(π/3) (cosec^2 2θ−1)dθ    =(1/2)∫_(π/6) ^(π/3) cosec^2 2θ−(1/2)∫_(π/6) ^(π/3) 1dθ    =−(1/4)[cot2θ]_(π/6) ^(π/3)  −(1/2)[θ]_(π/6) ^(π/3)     =−(1/4){cot(((2θ)/3))−cot((θ/3))}−(1/2)((π/3)−(π/6))    =−(1/4)(−(1/(√3))−(1/(√3)))−(1/2)((π/6))    =(1/(2(√3)))−(π/(12))

I=12π/6π/3(cosec22θ1)dθ=12π/6π/3cosec22θ12π/6π/31dθ=14[cot2θ]π/6π/312[θ]π/6π/3=14{cot(2θ3)cot(θ3)}12(π3π6)=14(1313)12(π6)=123π12

Commented by Isse last updated on 01/Oct/17

thanks

thanks

Commented by Tikufly last updated on 02/Oct/17

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