Menu Close

be-Math-cos-x-dx-sin-2-x-4sin-x-5-




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

Leave a Reply

Your email address will not be published. Required fields are marked *