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J-d-tan-cot-sec-csc-




Question Number 129173 by bramlexs22 last updated on 13/Jan/21
 J = ∫ (dθ/(tan θ+cot θ+sec θ+csc θ)) ?
J=dθtanθ+cotθ+secθ+cscθ?
Answered by liberty last updated on 13/Jan/21
 J = ∫(dθ/(((sin θ+1)/(cos θ))+((cos θ+1)/(sin θ)))) = ∫ ((cos θ sin θ dθ)/(sin^2 θ+sin θ+cos^2 θ+cos θ))   J = ∫ ((sin θ cos θ dθ)/(1+sin θ+cos θ)) = ∫ ((2sin ((θ/2))cos ((θ/2))cos θ)/(2sin ((θ/2))cos ((θ/2))+2cos^2 ((θ/2))))dθ  J = ∫ ((sin ((θ/2))cos θ)/(sin ((θ/2))+cos ((θ/2)))) dθ   J= ∫sin ((θ/2))(cos ((θ/2))−sin ((θ/2)))dθ   J = ∫ (1/2)sin θ−((1/2)−(1/2)cos θ)dθ   J =−(1/2)cos 𝛉−(1/2)θ+(1/2)sin 𝛉+C
J=dθsinθ+1cosθ+cosθ+1sinθ=cosθsinθdθsin2θ+sinθ+cos2θ+cosθJ=sinθcosθdθ1+sinθ+cosθ=2sin(θ2)cos(θ2)cosθ2sin(θ2)cos(θ2)+2cos2(θ2)dθJ=sin(θ2)cosθsin(θ2)+cos(θ2)dθJ=sin(θ2)(cos(θ2)sin(θ2))dθJ=12sinθ(1212cosθ)dθJ=12cosθ12θ+12sinθ+C

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