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Question Number 164612 by bounhome last updated on 19/Jan/22

solve: 1. ∫(1/(sinx))dx 2.∫(1/(cosx))dx

$${solve}: \\ $$$$\:\mathrm{1}.\:\int\frac{\mathrm{1}}{{sinx}}{dx} \\ $$$$\:\mathrm{2}.\int\frac{\mathrm{1}}{{cosx}}{dx} \\ $$

Answered by Ar Brandon last updated on 19/Jan/22

∫(1/(sinx))dx, x=2t⇒dx=2dt =∫((2dt)/(sin2t))=∫(1/(sintcost))dt=∫((cos^2 t+sin^2 t)/(sintcost))dt =∫(((cost)/(sint))+((sint)/(cost)))dt=ln(sint)−ln(cost)+C =ln∣tan((x/2))∣+C

$$\int\frac{\mathrm{1}}{\mathrm{sin}{x}}{dx},\:{x}=\mathrm{2}{t}\Rightarrow{dx}=\mathrm{2}{dt} \\ $$$$=\int\frac{\mathrm{2}{dt}}{\mathrm{sin2}{t}}=\int\frac{\mathrm{1}}{\mathrm{sin}{t}\mathrm{cos}{t}}{dt}=\int\frac{\mathrm{cos}^{\mathrm{2}} {t}+\mathrm{sin}^{\mathrm{2}} {t}}{\mathrm{sin}{t}\mathrm{cos}{t}}{dt} \\ $$$$=\int\left(\frac{\mathrm{cos}{t}}{\mathrm{sin}{t}}+\frac{\mathrm{sin}{t}}{\mathrm{cos}{t}}\right){dt}=\mathrm{ln}\left(\mathrm{sin}{t}\right)−\mathrm{ln}\left(\mathrm{cos}{t}\right)+{C} \\ $$$$=\mathrm{ln}\mid\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)\mid+{C} \\ $$

Answered by Ar Brandon last updated on 19/Jan/22

∫(1/(cosx))dx= Q163829

$$\int\frac{\mathrm{1}}{\mathrm{cos}{x}}{dx}=\:{Q}\mathrm{163829} \\ $$

Answered by Eulerian last updated on 20/Jan/22

Solution: 1. ∫ ((sin x)/(sin^2 x)) dx = ∫ ((sin x)/(1−cos^2 x)) = −tanh^(−1) (cos x) + C 2. ∫ ((cos x)/(cos^2 x)) dx = ∫ ((cos x)/(1−sin^2 x)) dx = tanh^(−1) (sin x) + C

$$\: \\ $$$$\:\boldsymbol{\mathrm{Solution}}: \\ $$$$\:\mathrm{1}.\:\int\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:=\:−\mathrm{tanh}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right)\:+\:\mathrm{C} \\ $$$$\: \\ $$$$\:\mathrm{2}.\:\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=\:\mathrm{tanh}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right)\:+\:\mathrm{C} \\ $$$$\: \\ $$

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