Question Number 71717 by peter frank last updated on 19/Oct/19

Commented bymathmax by abdo last updated on 19/Oct/19

let I=∫ (dx/(cos(x−a)cos(x−b))) ⇒I=∫ ((2dx)/(cos(2x−a−b)+cos(a−b))) =_(2x−(a+b)=t) ∫ (dt/(cost +λ)) with λ=cos(a−b) changement tan((t/2))=u give ∫ (dt/(cost +λ)) =∫ (1/(((1−u^2 )/(1+u^2 ))+λ))((2du)/(1+u^2 )) =∫ ((2du)/(1−u^2 +λ(1+u^2 ))) =∫ ((2du)/((λ−1)u^2 +1+λ)) =∫((−2du)/((1−λ)u^2 −(1+λ))) =((−2)/(1−λ)) ∫ (du/(u^2 −((1+λ)/(1−λ))))( look that ((1+λ)/(1−λ))>0) =_(u=(√((1+λ)/(1−λ)))z) ((−2)/(1−λ))((1−λ)/(1+λ)) ∫ (1/(z^2 −1))((√(1+λ))/(√(1−λ)))dz =((−1)/(√(1−λ^2 ))) ∫((1/(z−1))−(1/(z+1)))dz =((−1)/(√(1−λ^2 )))ln∣((z−1)/(z+1))∣ +C =((−1)/(√(1−cos^2 (a−b)))) ×ln∣(((√((1−λ)/(1+λ)))tan((t/2))−1)/((√((1−λ)/(1+λ)))tan((t/2))+1))∣ +C =((−1)/(∣sin(a−b)∣))×ln∣((tan(((a−b)/2))tan(x−((a+b)/2))−1)/(tan(((a−b)/2))tan(x−((a+b)/2))+1))∣ +C with a−b≠kπ k∈Z

Commented bypeter frank last updated on 20/Oct/19

thank you

Commented bymathmax by abdo last updated on 23/Oct/19

you are welcome.

Answered by Tanmay chaudhury last updated on 19/Oct/19

(1/(sin(a−b)))∫((sin{(x−b)−(x−a)})/(cos(x−a)cos(x−b)))dx (1/(sin(a−b)))∫((sin(x−b)cos(x−a)−cos(x−b)sin(x−a))/(cos(x−a)cos(x−b)))dx (1/(sin(a−b)))∫[tan(x−b)−tan(x−a)]dx (1/(sin(a−b)))[lnsec(x−b)−lnsec(x−a)]+c (1/(sin(a−b)))ln(((sec(x−b))/(sec(x−a))))+c

Commented by$@ty@m123 last updated on 19/Oct/19

Quite easier but tricky.

Commented bypeter frank last updated on 19/Oct/19

thank you very much

Commented byPrithwish sen last updated on 19/Oct/19

tanmoy sir pujo kamon katlo. subho bijaya. kothae chilen ? bhalo achen ?