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Question-84557




Question Number 84557 by Power last updated on 14/Mar/20
Commented by jagoll last updated on 14/Mar/20
ln(cos x) = u ⇒ cos x = e^u   −sin x dx = e^u  du   dx = ((−e^u )/(sin x)) dx = ((−e^u  du)/( (√(1−e^(2u) ))))  ∫ ((−e^(2u) )/( (√(1−e^(2u) )))) du = ∫ ((d(1−e^(2u) ))/( (√(1−e^(2u) ))))  = 2(√(1−e^(2u) )) + c  = 2(√(1−cos^2 x)) +c  = 2 ∣sin x∣ + c
$$\mathrm{ln}\left(\mathrm{cos}\:\mathrm{x}\right)\:=\:\mathrm{u}\:\Rightarrow\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{e}^{\mathrm{u}} \\ $$$$−\mathrm{sin}\:\mathrm{x}\:\mathrm{dx}\:=\:\mathrm{e}^{\mathrm{u}} \:\mathrm{du}\: \\ $$$$\mathrm{dx}\:=\:\frac{−\mathrm{e}^{\mathrm{u}} }{\mathrm{sin}\:\mathrm{x}}\:\mathrm{dx}\:=\:\frac{−\mathrm{e}^{\mathrm{u}} \:\mathrm{du}}{\:\sqrt{\mathrm{1}−\mathrm{e}^{\mathrm{2u}} }} \\ $$$$\int\:\frac{−\mathrm{e}^{\mathrm{2u}} }{\:\sqrt{\mathrm{1}−\mathrm{e}^{\mathrm{2u}} }}\:\mathrm{du}\:=\:\int\:\frac{\mathrm{d}\left(\mathrm{1}−\mathrm{e}^{\mathrm{2u}} \right)}{\:\sqrt{\mathrm{1}−\mathrm{e}^{\mathrm{2u}} }} \\ $$$$=\:\mathrm{2}\sqrt{\mathrm{1}−\mathrm{e}^{\mathrm{2u}} }\:+\:\mathrm{c} \\ $$$$=\:\mathrm{2}\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:+\mathrm{c} \\ $$$$=\:\mathrm{2}\:\mid\mathrm{sin}\:\mathrm{x}\mid\:+\:\mathrm{c}\: \\ $$

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