Question Number 147593 by 0731619 last updated on 22/Jul/21
Answered by gsk2684 last updated on 22/Jul/21
$${put}\:\mathrm{sin}\:{y}\:=\:{t}\Rightarrow\mathrm{cos}\:{y}\:{dy}\:=\:{dt} \\ $$$$\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} +{t}−\mathrm{6}}{dt}=\int\frac{\mathrm{1}}{\left({t}−\mathrm{2}\right)\left({t}+\mathrm{3}\right)}{dt} \\ $$$$=\int\left(\frac{\frac{\mathrm{1}}{\mathrm{5}}}{{t}−\mathrm{2}}+\frac{−\frac{\mathrm{1}}{\mathrm{5}}}{{t}+\mathrm{3}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{ln}\:\left({t}−\mathrm{2}\right)−\mathrm{ln}\:\left({t}+\mathrm{3}\right)\right)+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\left(\frac{{t}−\mathrm{2}}{{t}+\mathrm{3}}\right)+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\left(\frac{\mathrm{sin}\:{y}−\mathrm{2}}{\mathrm{sin}\:{y}+\mathrm{3}}\right)+{c} \\ $$
Commented by 0731619 last updated on 22/Jul/21
$${thanks} \\ $$
Answered by Olaf_Thorendsen last updated on 22/Jul/21
$$ \\ $$$${f}\left({y}\right)\:=\:\frac{\mathrm{cos}{y}}{\mathrm{sin}^{\mathrm{2}} {y}+\mathrm{sin}{y}−\mathrm{6}} \\ $$$${f}\left({y}\right)\:=\:\frac{\mathrm{cos}{y}}{\left(\mathrm{sin}{y}−\mathrm{2}\right)\left(\mathrm{sin}{y}+\mathrm{3}\right)} \\ $$$${f}\left({y}\right)\:=\:\frac{\frac{\mathrm{cos}{y}}{\left(\mathrm{sin}{y}+\mathrm{3}\right)^{\mathrm{2}} }}{\left(\frac{\mathrm{sin}{y}−\mathrm{2}}{\mathrm{sin}{y}+\mathrm{3}}\right)} \\ $$$${f}\left({y}\right)\:=\:\frac{\mathrm{1}}{\mathrm{5}}.\frac{\frac{\mathrm{cos}{y}\left(\mathrm{sin}{y}+\mathrm{3}\right)−\left(\mathrm{sin}{y}−\mathrm{2}\right)\mathrm{cos}{y}}{\left(\mathrm{sin}{y}+\mathrm{3}\right)^{\mathrm{2}} }}{\left(\frac{\mathrm{sin}{y}−\mathrm{2}}{\mathrm{sin}{y}+\mathrm{3}}\right)} \\ $$$${f}\left({y}\right)\:=\frac{\mathrm{1}}{\mathrm{5}}.\frac{\frac{{d}}{{dy}}\left(\frac{\mathrm{sin}{y}−\mathrm{2}}{\mathrm{sin}{y}+\mathrm{3}}\right)}{\left(\frac{\mathrm{sin}{y}−\mathrm{2}}{\mathrm{sin}{y}+\mathrm{3}}\right)} \\ $$$$\Rightarrow\:\mathrm{F}\left({y}\right)\:=\:\int{f}\left({y}\right){dy}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\mid\frac{\mathrm{sin}{y}−\mathrm{2}}{\mathrm{sin}{y}+\mathrm{3}}\mid+\mathrm{C} \\ $$
Commented by 0731619 last updated on 22/Jul/21
$${thanks} \\ $$$$ \\ $$