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Question-139393

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Question Number 139393 by BHOOPENDRA last updated on 26/Apr/21
Answered by Ar Brandon last updated on 26/Apr/21
I=∫((cosx)/(2cosx+sinx+3))dx, t=tan(x/2)⇒dt=((1+t^2 )/2)dx =∫(((1−t^2 )/(1+t^2 ))/(2(((1−t^2 )/(1+t^2 )))+((2t)/(1+t^2 ))+3))∙(2/(1+t^2 ))dt =∫((2(1−t^2 ))/(2(1−t^2 )+2t+3(1+t^2 )))∙(dt/(1+t^2 )) =∫((2(1−t^2 ))/((5+2t+t^2 )(1+t^2 )))dt f(t)=((2(1−t^2 ))/((5+2t+t^2 )(1+t^2 )))=((at+b)/(t^2 +2t+5))+((ct+d)/(t^2 +1)) =(((at+b)(t^2 +1)+(t^2 +2t+5)(ct+d))/((5+2t+t^2 )(1+t^2 ))) a+c=0, b+d+2c=−2, a+2d+5c=0, b+5d=2 ⇒b=−2, d=(4/5), c=−(2/5), a=(2/5) I=∫(((2/5)t−2)/(t^2 +2t+5))dt−∫(((2/5)t−(4/5))/(t^2 +1))dt
$$\mathrm{I}=\int\frac{\mathrm{cos}{x}}{\mathrm{2cos}{x}+\mathrm{sin}{x}+\mathrm{3}}{dx},\:{t}=\mathrm{tan}\frac{{x}}{\mathrm{2}}\Rightarrow{dt}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}}{dx} \\ $$$$\:\:=\int\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{2}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{3}}\centerdot\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\:\:=\int\frac{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+\mathrm{2}{t}+\mathrm{3}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\centerdot\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\: \\ $$$$\:\:=\int\frac{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{5}+\mathrm{2}{t}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$${f}\left({t}\right)=\frac{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{5}+\mathrm{2}{t}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\frac{{at}+{b}}{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{5}}+\frac{{ct}+{d}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\left({at}+{b}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)+\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{5}\right)\left({ct}+{d}\right)}{\left(\mathrm{5}+\mathrm{2}{t}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$${a}+{c}=\mathrm{0},\:{b}+{d}+\mathrm{2}{c}=−\mathrm{2},\:{a}+\mathrm{2}{d}+\mathrm{5}{c}=\mathrm{0},\:{b}+\mathrm{5}{d}=\mathrm{2} \\ $$$$\Rightarrow{b}=−\mathrm{2},\:{d}=\frac{\mathrm{4}}{\mathrm{5}},\:{c}=−\frac{\mathrm{2}}{\mathrm{5}},\:{a}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\mathrm{I}=\int\frac{\frac{\mathrm{2}}{\mathrm{5}}{t}−\mathrm{2}}{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{5}}{dt}−\int\frac{\frac{\mathrm{2}}{\mathrm{5}}{t}−\frac{\mathrm{4}}{\mathrm{5}}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$

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