Question-147593

Question Number 147593 by 0731619 last updated on 22/Jul/21

Answered by gsk2684 last updated on 22/Jul/21

put sin y = t⇒cos y dy = dt ∫(1/(t^2 +t−6))dt=∫(1/((t−2)(t+3)))dt =∫(((1/5)/(t−2))+((−(1/5))/(t+3)))dt =(1/5)(ln (t−2)−ln (t+3))+c =(1/5)ln (((t−2)/(t+3)))+c =(1/5)ln (((sin y−2)/(sin y+3)))+c

$${put}\:\mathrm{sin}\:{y}\:=\:{t}\Rightarrow\mathrm{cos}\:{y}\:{dy}\:=\:{dt} \\ $$$$\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} +{t}−\mathrm{6}}{dt}=\int\frac{\mathrm{1}}{\left({t}−\mathrm{2}\right)\left({t}+\mathrm{3}\right)}{dt} \\ $$$$=\int\left(\frac{\frac{\mathrm{1}}{\mathrm{5}}}{{t}−\mathrm{2}}+\frac{−\frac{\mathrm{1}}{\mathrm{5}}}{{t}+\mathrm{3}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{ln}\:\left({t}−\mathrm{2}\right)−\mathrm{ln}\:\left({t}+\mathrm{3}\right)\right)+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\left(\frac{{t}−\mathrm{2}}{{t}+\mathrm{3}}\right)+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\left(\frac{\mathrm{sin}\:{y}−\mathrm{2}}{\mathrm{sin}\:{y}+\mathrm{3}}\right)+{c} \\ $$

Commented by 0731619 last updated on 22/Jul/21

$${thanks} \\ $$

Answered by Olaf_Thorendsen last updated on 22/Jul/21

f(y) = ((cosy)/(sin^2 y+siny−6)) f(y) = ((cosy)/((siny−2)(siny+3))) f(y) = (((cosy)/((siny+3)^2 ))/((((siny−2)/(siny+3))))) f(y) = (1/5).(((cosy(siny+3)−(siny−2)cosy)/((siny+3)^2 ))/((((siny−2)/(siny+3))))) f(y) =(1/5).(((d/dy)(((siny−2)/(siny+3))))/((((siny−2)/(siny+3))))) ⇒ F(y) = ∫f(y)dy = (1/5)ln∣((siny−2)/(siny+3))∣+C

$$ \\ $$$${f}\left({y}\right)\:=\:\frac{\mathrm{cos}{y}}{\mathrm{sin}^{\mathrm{2}} {y}+\mathrm{sin}{y}−\mathrm{6}} \\ $$$${f}\left({y}\right)\:=\:\frac{\mathrm{cos}{y}}{\left(\mathrm{sin}{y}−\mathrm{2}\right)\left(\mathrm{sin}{y}+\mathrm{3}\right)} \\ $$$${f}\left({y}\right)\:=\:\frac{\frac{\mathrm{cos}{y}}{\left(\mathrm{sin}{y}+\mathrm{3}\right)^{\mathrm{2}} }}{\left(\frac{\mathrm{sin}{y}−\mathrm{2}}{\mathrm{sin}{y}+\mathrm{3}}\right)} \\ $$$${f}\left({y}\right)\:=\:\frac{\mathrm{1}}{\mathrm{5}}.\frac{\frac{\mathrm{cos}{y}\left(\mathrm{sin}{y}+\mathrm{3}\right)−\left(\mathrm{sin}{y}−\mathrm{2}\right)\mathrm{cos}{y}}{\left(\mathrm{sin}{y}+\mathrm{3}\right)^{\mathrm{2}} }}{\left(\frac{\mathrm{sin}{y}−\mathrm{2}}{\mathrm{sin}{y}+\mathrm{3}}\right)} \\ $$$${f}\left({y}\right)\:=\frac{\mathrm{1}}{\mathrm{5}}.\frac{\frac{{d}}{{dy}}\left(\frac{\mathrm{sin}{y}−\mathrm{2}}{\mathrm{sin}{y}+\mathrm{3}}\right)}{\left(\frac{\mathrm{sin}{y}−\mathrm{2}}{\mathrm{sin}{y}+\mathrm{3}}\right)} \\ $$$$\Rightarrow\:\mathrm{F}\left({y}\right)\:=\:\int{f}\left({y}\right){dy}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\mid\frac{\mathrm{sin}{y}−\mathrm{2}}{\mathrm{sin}{y}+\mathrm{3}}\mid+\mathrm{C} \\ $$

Commented by 0731619 last updated on 22/Jul/21

$${thanks} \\ $$$$ \\ $$

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