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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)) ?
$$\:\mathrm{J}\:=\:\int\:\frac{\mathrm{d}\theta}{\mathrm{tan}\:\theta+\mathrm{cot}\:\theta+\mathrm{sec}\:\theta+\mathrm{csc}\:\theta}\:? \\ $$
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
$$\:\mathrm{J}\:=\:\int\frac{\mathrm{d}\theta}{\frac{\mathrm{sin}\:\theta+\mathrm{1}}{\mathrm{cos}\:\theta}+\frac{\mathrm{cos}\:\theta+\mathrm{1}}{\mathrm{sin}\:\theta}}\:=\:\int\:\frac{\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\:\mathrm{d}\theta}{\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{sin}\:\theta+\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{cos}\:\theta} \\ $$$$\:\mathrm{J}\:=\:\int\:\frac{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:\mathrm{d}\theta}{\mathrm{1}+\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}\:=\:\int\:\frac{\mathrm{2sin}\:\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\:\theta}{\mathrm{2sin}\:\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\theta}{\mathrm{2}}\right)+\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}\mathrm{d}\theta \\ $$$$\mathrm{J}\:=\:\int\:\frac{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\:\theta}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\theta}{\mathrm{2}}\right)}\:\mathrm{d}\theta \\ $$$$\:\mathrm{J}=\:\int\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}\right)\left(\mathrm{cos}\:\left(\frac{\theta}{\mathrm{2}}\right)−\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}\right)\right)\mathrm{d}\theta \\ $$$$\:\mathrm{J}\:=\:\int\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta−\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\theta\right)\mathrm{d}\theta \\ $$$$\:\underline{\boldsymbol{\mathrm{J}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\:\boldsymbol{\theta}−\frac{\mathrm{1}}{\mathrm{2}}\theta+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{sin}}\:\boldsymbol{\theta}+\boldsymbol{\mathrm{C}}\:} \\ $$$$\: \\ $$

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