Question Number 109872 by bemath last updated on 26/Aug/20
$$\:\:\frac{\bigstar{be}\bigstar}{\mathcal{M}{ath}} \\ $$$$\int\:\frac{\mathrm{cos}\:{x}\:{dx}}{\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{4sin}\:{x}−\mathrm{5}}\:? \\ $$
Commented by bemath last updated on 26/Aug/20
$${santuyy}\:{all}\:{master} \\ $$
Answered by Sarah85 last updated on 26/Aug/20
$${t}=\mathrm{sin}\:{x} \\ $$$$\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{4}{t}−\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{5}}\:... \\ $$
Answered by 1549442205PVT last updated on 26/Aug/20
$$\mathrm{Put}\:\mathrm{sin}\:\mathrm{x}=\mathrm{u}\Rightarrow\mathrm{du}=\mathrm{coxdxdx} \\ $$$$\mathrm{F}=\int\:\frac{\mathrm{cos}\:{x}\:{dx}}{\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{4sin}\:{x}−\mathrm{5}}=\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{4u}−\mathrm{5}} \\ $$$$=\int\frac{\mathrm{du}}{\left(\mathrm{u}−\mathrm{1}\right)\left(\mathrm{u}+\mathrm{5}\right)}=\frac{\mathrm{1}}{\mathrm{6}}\int\left(\frac{\mathrm{1}}{\mathrm{u}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{u}+\mathrm{5}}\right)\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{ln}\mid\mathrm{u}−\mathrm{1}\mid−\mathrm{ln}\mid\mathrm{u}+\mathrm{5}\mid\right)=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\mid\frac{\mathrm{u}−\mathrm{1}}{\mathrm{u}+\mathrm{5}}\mid+\mathrm{C} \\ $$$$\boldsymbol{\mathrm{F}}=\frac{\mathrm{1}}{\mathrm{6}}\boldsymbol{\mathrm{ln}}\mid\frac{\boldsymbol{\mathrm{sinx}}−\mathrm{1}}{\boldsymbol{\mathrm{sinx}}+\mathrm{5}}\mid+\boldsymbol{\mathrm{C}} \\ $$