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Question Number 79086 by key of knowledge last updated on 22/Jan/20

if:∫cos(f(x))dx=g(x) ∫sin(f(x))dx=? (use g(x))

$$\mathrm{if}:\int\mathrm{cos}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\mathrm{dx}=\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\int\mathrm{sin}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\mathrm{dx}=?\:\left(\mathrm{use}\:\mathrm{g}\left(\mathrm{x}\right)\right) \\ $$

Commented by mr W last updated on 23/Jan/20

how did you get ∫ (du/(−f′(x)))= −((cos (f(x)))/(−f′(x)))+c

$${how}\:{did}\:{you}\:{get} \\ $$$$\int\:\frac{\mathrm{du}}{−\mathrm{f}'\left(\mathrm{x}\right)}=\:−\frac{\mathrm{cos}\:\left(\mathrm{f}\left(\mathrm{x}\right)\right)}{−\mathrm{f}'\left(\mathrm{x}\right)}+\mathrm{c} \\ $$

Commented by jagoll last updated on 23/Jan/20

cos f(x)=g′(x)=u −f′(x)sin (f(x))dx=du sin f(x)=(√(1−cos^2 (f(x)))) sin f(x)=(√(1−u^2 )) ∫(√(1−u^2 )) ×(du/(−f′(x).(√(1−u^2 )))) = ∫ (du/(−f′(x)))= −((cos (f(x)))/(−f′(x)))+c

$$\mathrm{cos}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}'\left(\mathrm{x}\right)=\mathrm{u} \\ $$$$−\mathrm{f}'\left(\mathrm{x}\right)\mathrm{sin}\:\left(\mathrm{f}\left(\mathrm{x}\right)\right)\mathrm{dx}=\mathrm{du} \\ $$$$\mathrm{sin}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{f}\left(\mathrm{x}\right)\right)} \\ $$$$\mathrm{sin}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} } \\ $$$$\int\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }\:×\frac{\mathrm{du}}{−\mathrm{f}'\left(\mathrm{x}\right).\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }}\:= \\ $$$$\int\:\frac{\mathrm{du}}{−\mathrm{f}'\left(\mathrm{x}\right)}=\:−\frac{\mathrm{cos}\:\left(\mathrm{f}\left(\mathrm{x}\right)\right)}{−\mathrm{f}'\left(\mathrm{x}\right)}+\mathrm{c} \\ $$

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