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Question Number 153611 by saly last updated on 08/Sep/21

Answered by Ar Brandon last updated on 08/Sep/21

A=∫((3sinx+2cosx)/(3cosx+2sinx))dx =∫((2cosx−3sinx)/(2sinx+3cosx))dx+6∫((sinx)/(2sinx+3cosx))dx =ln∣2sinx+3cosx∣+6∫(((4t)/(1+t^2 ))/(((4t)/(1+t^2 ))+3((1−t^2 )/(1+t^2 ))))∙(dt/(1+t^2 )) =ln∣2sinx+3cosx∣+6∫((4tdt)/((3+4t−3t^2 )(1+t^2 ))) You may proceed with the decomposition ((4t)/((3t^2 −4t−3)(t^2 +1)))=((at+b)/(3t^2 −4t−3))+((ct+d)/(t^2 +1))

$${A}=\int\frac{\mathrm{3sin}{x}+\mathrm{2cos}{x}}{\mathrm{3cos}{x}+\mathrm{2sin}{x}}{dx} \\ $$$$\:\:\:\:=\int\frac{\mathrm{2cos}{x}−\mathrm{3sin}{x}}{\mathrm{2sin}{x}+\mathrm{3cos}{x}}{dx}+\mathrm{6}\int\frac{\mathrm{sin}{x}}{\mathrm{2sin}{x}+\mathrm{3cos}{x}}{dx} \\ $$$$\:\:\:\:=\mathrm{ln}\mid\mathrm{2sin}{x}+\mathrm{3cos}{x}\mid+\mathrm{6}\int\frac{\frac{\mathrm{4}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{4}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{3}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\centerdot\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\mathrm{ln}\mid\mathrm{2sin}{x}+\mathrm{3cos}{x}\mid+\mathrm{6}\int\frac{\mathrm{4}{tdt}}{\left(\mathrm{3}+\mathrm{4}{t}−\mathrm{3}{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$\mathrm{You}\:\mathrm{may}\:\mathrm{proceed}\:\mathrm{with}\:\mathrm{the}\:\mathrm{decomposition} \\ $$$$\frac{\mathrm{4}{t}}{\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{3}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{at}+{b}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{3}}+\frac{{ct}+{d}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$

Commented by saly last updated on 08/Sep/21

Thank you But in number B.C.D ?

$$\:\:{Thank}\:{you} \\ $$$$\:\:\:{But}\:{in}\:{number}\:\:{B}.{C}.{D}\:? \\ $$

Answered by Ar Brandon last updated on 08/Sep/21

B=∫arctan((√(1−x^2 )))dx { ((u(x)=arctan((√(1−x^2 ))))),((v′(x)=1)) :}⇒ { ((u′(x)=−(x/( (√(1−x^2 ))))∙(1/(2−x^2 )))),((v(x)=x)) :} B=xarctan((√(1−x^2 )))+∫((x^2 dx)/((2−x^2 )(√(1−x^2 )))) =xarctan((√(1−x^2 )))−∫(dx/( (√(1−x^2 ))))+2∫(dx/((2−x^2 )(√(1−x^2 )))) =xarctan((√(1−x^2 )))−sin^(−1) (x)+2∫(dx/(x^3 ((2/x^2 )−1)(√((1/x^2 )−1)))) I=∫(dx/(x^3 ((2/x^2 )−1)(√((1/x^2 )−1)))), u=(1/x^2 )⇒du=−(2/x^3 )dx =−(1/2)∫(du/((2u−1)(√(u−1)))), t^2 =u−1⇒2t=du =−∫((tdt)/((2t^2 +1)t))=−(1/( (√2)))tan^(−1) ((√2)t)+C B=xarctan((√(1−x^2 )))−sin^(−1) (x)−(√2)tan^(−1) ((((√2)∙(√(1−x^2 )))/x))+C

$${B}=\int\mathrm{arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$$$\begin{cases}{{u}\left({x}\right)=\mathrm{arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}\\{{v}'\left({x}\right)=\mathrm{1}}\end{cases}\Rightarrow\begin{cases}{{u}'\left({x}\right)=−\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\centerdot\frac{\mathrm{1}}{\mathrm{2}−{x}^{\mathrm{2}} }}\\{{v}\left({x}\right)={x}}\end{cases} \\ $$$${B}={x}\mathrm{arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)+\int\frac{{x}^{\mathrm{2}} {dx}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\:\:\:\:={x}\mathrm{arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+\mathrm{2}\int\frac{{dx}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\:\:\:\:={x}\mathrm{arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\mathrm{sin}^{−\mathrm{1}} \left({x}\right)+\mathrm{2}\int\frac{{dx}}{{x}^{\mathrm{3}} \left(\frac{\mathrm{2}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}}} \\ $$$${I}=\int\frac{{dx}}{{x}^{\mathrm{3}} \left(\frac{\mathrm{2}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}}},\:{u}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\Rightarrow{du}=−\frac{\mathrm{2}}{{x}^{\mathrm{3}} }{dx} \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{\left(\mathrm{2}{u}−\mathrm{1}\right)\sqrt{{u}−\mathrm{1}}},\:{t}^{\mathrm{2}} ={u}−\mathrm{1}\Rightarrow\mathrm{2}{t}={du} \\ $$$$\:\:=−\int\frac{{tdt}}{\left(\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right){t}}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}{t}\right)+{C} \\ $$$${B}={x}\mathrm{arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\mathrm{sin}^{−\mathrm{1}} \left({x}\right)−\sqrt{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\centerdot\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\right)+{C} \\ $$

Answered by Ar Brandon last updated on 08/Sep/21

C=∫(dx/( (√(2x^2 −8x+20))))=(1/( (√2)))∫(dx/( (√(x^2 −4x+10)))) =(1/( (√2)))∫(dx/( (√((x−2)^2 +6))))=(1/( (√2)))arcsin(((x−2)/( (√6))))+K, K∈R

$${C}=\int\frac{{dx}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{20}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{10}}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dx}}{\:\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{6}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{arcsin}\left(\frac{{x}−\mathrm{2}}{\:\sqrt{\mathrm{6}}}\right)+{K},\:{K}\in\mathbb{R} \\ $$

Commented by saly last updated on 08/Sep/21

Thank you very much

$$\:\:{Thank}\:\:{you}\:{very}\:{much} \\ $$

Commented by Ar Brandon last updated on 08/Sep/21

If D was ∫((x^4 dx)/((x^5 +1)(√(5−x^5 )))) it would have been ∫((x^4 dx)/((x^5 +1)(√(5−x^5 )))), u=x^5 ⇒du=5x^4 dx =(1/5)∫(du/((u+1)(√(5−u)))), 5+t^2 =u⇒2tdt=du =(2/5)∫((tdt)/((t^2 +6)t))=(2/(5(√6)))tan^(−1) ((t/( (√6))))+C =((√6)/(15))tan^(−1) ((√((5−x^5 )/6)))+C

$$\mathrm{If}\:{D}\:\mathrm{was}\:\int\frac{{x}^{\mathrm{4}} {dx}}{\left({x}^{\mathrm{5}} +\mathrm{1}\right)\sqrt{\mathrm{5}−{x}^{\mathrm{5}} }}\:\mathrm{it}\:\mathrm{would}\:\mathrm{have}\:\mathrm{been} \\ $$$$\int\frac{{x}^{\mathrm{4}} {dx}}{\left({x}^{\mathrm{5}} +\mathrm{1}\right)\sqrt{\mathrm{5}−{x}^{\mathrm{5}} }},\:{u}={x}^{\mathrm{5}} \Rightarrow{du}=\mathrm{5}{x}^{\mathrm{4}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{du}}{\left({u}+\mathrm{1}\right)\sqrt{\mathrm{5}−{u}}},\:\mathrm{5}+{t}^{\mathrm{2}} ={u}\Rightarrow\mathrm{2}{tdt}={du} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\int\frac{{tdt}}{\left({t}^{\mathrm{2}} +\mathrm{6}\right){t}}=\frac{\mathrm{2}}{\mathrm{5}\sqrt{\mathrm{6}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{\mathrm{6}}}\right)+{C} \\ $$$$=\frac{\sqrt{\mathrm{6}}}{\mathrm{15}}\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{5}−{x}^{\mathrm{5}} }{\mathrm{6}}}\right)+{C} \\ $$

Commented by Ar Brandon last updated on 08/Sep/21

You're welcome

Commented by puissant last updated on 08/Sep/21

sacré Brandon ������������

Commented by Ar Brandon last updated on 08/Sep/21

Bonsoir bro

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