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Question Number 49838 by MJS last updated on 11/Dec/18

∫(dx/(√((a+1)cos 2x +4cos x −a+3)))=?

$$\int\frac{{dx}}{\sqrt{\left({a}+\mathrm{1}\right)\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{4cos}\:{x}\:−{a}+\mathrm{3}}}=? \\ $$

Commented by MJS last updated on 11/Dec/18

∫(dx/(√((a+1)cos 2x +4cos x −a+3)))=       [t=(x/2) → dx=2dt]  =2∫(dt/(√((a+1)cos 4t +4cos 2t −a+3)))=  =((√2)/2)∫(dt/(cos t (√((a+1)cos^2  t −a))))=       [u=tan t → dt=(du/(u^2 +1)); cos t =(1/(√(u^2 +1)))]  =((√2)/2)∫(du/(√(1−au^2 )))=       [v=(√a)u → du=(dv/(√a))]  =(1/(√(2a)))∫(dv/(√(1−v^2 )))=(1/(√(2a)))arcsin v =(1/(√(2a)))arcsin ((√a)u) =  =(1/(√(2a)))arcsin ((√a)tan t) =(1/(√(2a)))arcsin ((√a)tan (x/2)) +C

$$\int\frac{{dx}}{\sqrt{\left({a}+\mathrm{1}\right)\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{4cos}\:{x}\:−{a}+\mathrm{3}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\mathrm{2}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{\sqrt{\left({a}+\mathrm{1}\right)\mathrm{cos}\:\mathrm{4}{t}\:+\mathrm{4cos}\:\mathrm{2}{t}\:−{a}+\mathrm{3}}}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{cos}\:{t}\:\sqrt{\left({a}+\mathrm{1}\right)\mathrm{cos}^{\mathrm{2}} \:{t}\:−{a}}}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{tan}\:{t}\:\rightarrow\:{dt}=\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}};\:\mathrm{cos}\:{t}\:=\frac{\mathrm{1}}{\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}\right] \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{du}}{\sqrt{\mathrm{1}−{au}^{\mathrm{2}} }}= \\ $$$$\:\:\:\:\:\left[{v}=\sqrt{{a}}{u}\:\rightarrow\:{du}=\frac{{dv}}{\sqrt{{a}}}\right] \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}{a}}}\int\frac{{dv}}{\sqrt{\mathrm{1}−{v}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}{a}}}\mathrm{arcsin}\:{v}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}{a}}}\mathrm{arcsin}\:\left(\sqrt{{a}}{u}\right)\:= \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}{a}}}\mathrm{arcsin}\:\left(\sqrt{{a}}\mathrm{tan}\:{t}\right)\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}{a}}}\mathrm{arcsin}\:\left(\sqrt{{a}}\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\:+{C} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

∫(dx/(√(−a(1−cos2x)+cos2x+4cosx+3)))  ∫(dx/(√(4cosx+4−(1−cos2x)−a(1−cos2x))))  ∫(dx/(√(4(2cos^2 (x/2))−(2sin^2 x)−a(2sin^2 x))))  ∫(dx/(√(8cos^2 (x/2)−2(a+1)×4sin^2 (x/2)cos^2 (x/2))))  ∫((sec(x/2)dx)/(√(8−8(a+1)sin^2 (x/2))))  (1/(√8))×∫((sec(x/2)dx)/(√(1−(a+1)sin^2 (x/2))))  (1/((√(8(a+1))) ))∫((sec(x/2)dx)/(√((1/(a+1))−sin^2 (x/2))))  (1/(√(8(a+1))))∫((cos(x/2)dx)/(cos^2 (x/2)(√((1/(a+1))−sin^2 (x/2)))))dx  t=sin(x/2)   dt=(1/2)cos(x/2)dx  (1/(√(8(a+1))))∫((2dt)/((1−t^2 )(√((1/(a+1))−t^2 ))))  t=(1/k)   dt=((−1)/k^2 )dk  (1/(√(2(a+1))))∫((−dk)/(k^2 (1−(1/k^2 ))(√((1/(a+1))−(1/k^2 )))))  (1/(√(2(a+1))))∫((−kdk)/((k^2 −1)(√((k^2 /(a+1))−1))))  ((−1)/(√(2(a+1))))∫((kdk)/((k^2 −1)(√((k^2 /(a+1))−1))))  p=k^2  dp=2kdk  ((−1)/(√(2(a+1))))∫(dp/(2(p−1)(√((p/(a+1))−1))))  ((−1)/((√(8(a+1))) ))∫(dp/((p−1)(√((p/(a+1))−1))))  y^2 =(p/(a+1))−1    2ydy=(dp/(a+1))  =((−1)/(√(8(a+1))))∫((2(a+1)ydy)/({(a+1)(y^2 +1)−1}×y))  =((−2(a+1))/(√(8(a+1))))∫(dy/({(a+1)y^2 +a}))  =((−((√(a+1)) ))/(√2))×(1/(a+1))∫(dy/(y^2 +(a/(a+1))))  =((−1)/(√(2(a+1))))×(1/(√(a/(a+1))))tan^(−1) ((y/(√(a/(a+1)))))+c  ((−1)/(√(2a)))tan^(−1) ((((√((p/(a+1))−1)) )/(√(a/(a+1)))))+c  ((−1)/(√(2a)))tan^(−1) (((√((k^2 /(a+1))−1))/(√(a/(a+1)))))  =((−1)/(√(2a)))tan^(−1) ((((√((1/(t^2 (a+1)))−1)) )/(√(a/(a+1)))))  =((−1)/(√(2a)))tan^(−1) (((√((1/(sin^2 ((x/2))(a+1)))−1))/(√(a/(a+1)))))+c

$$\int\frac{{dx}}{\sqrt{−{a}\left(\mathrm{1}−{cos}\mathrm{2}{x}\right)+{cos}\mathrm{2}{x}+\mathrm{4}{cosx}+\mathrm{3}}} \\ $$$$\int\frac{{dx}}{\sqrt{\mathrm{4}{cosx}+\mathrm{4}−\left(\mathrm{1}−{cos}\mathrm{2}{x}\right)−{a}\left(\mathrm{1}−{cos}\mathrm{2}{x}\right)}} \\ $$$$\int\frac{{dx}}{\sqrt{\mathrm{4}\left(\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)−\left(\mathrm{2}{sin}^{\mathrm{2}} {x}\right)−{a}\left(\mathrm{2}{sin}^{\mathrm{2}} {x}\right)}} \\ $$$$\int\frac{{dx}}{\sqrt{\mathrm{8}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{2}\left({a}+\mathrm{1}\right)×\mathrm{4}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}} \\ $$$$\int\frac{{sec}\frac{{x}}{\mathrm{2}}{dx}}{\sqrt{\mathrm{8}−\mathrm{8}\left({a}+\mathrm{1}\right){sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{8}}}×\int\frac{{sec}\frac{{x}}{\mathrm{2}}{dx}}{\sqrt{\mathrm{1}−\left({a}+\mathrm{1}\right){sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{8}\left({a}+\mathrm{1}\right)}\:}\int\frac{{sec}\frac{{x}}{\mathrm{2}}{dx}}{\sqrt{\frac{\mathrm{1}}{{a}+\mathrm{1}}−{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{8}\left({a}+\mathrm{1}\right)}}\int\frac{{cos}\frac{{x}}{\mathrm{2}}{dx}}{{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{{a}+\mathrm{1}}−{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}}{dx} \\ $$$${t}={sin}\frac{{x}}{\mathrm{2}}\:\:\:{dt}=\frac{\mathrm{1}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}{dx} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{8}\left({a}+\mathrm{1}\right)}}\int\frac{\mathrm{2}{dt}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\sqrt{\frac{\mathrm{1}}{{a}+\mathrm{1}}−{t}^{\mathrm{2}} }} \\ $$$${t}=\frac{\mathrm{1}}{{k}}\:\:\:{dt}=\frac{−\mathrm{1}}{{k}^{\mathrm{2}} }{dk} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}\left({a}+\mathrm{1}\right)}}\int\frac{−{dk}}{{k}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\sqrt{\frac{\mathrm{1}}{{a}+\mathrm{1}}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}\left({a}+\mathrm{1}\right)}}\int\frac{−{kdk}}{\left({k}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\frac{{k}^{\mathrm{2}} }{{a}+\mathrm{1}}−\mathrm{1}}} \\ $$$$\frac{−\mathrm{1}}{\sqrt{\mathrm{2}\left({a}+\mathrm{1}\right)}}\int\frac{{kdk}}{\left({k}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\frac{{k}^{\mathrm{2}} }{{a}+\mathrm{1}}−\mathrm{1}}} \\ $$$${p}={k}^{\mathrm{2}} \:{dp}=\mathrm{2}{kdk} \\ $$$$\frac{−\mathrm{1}}{\sqrt{\mathrm{2}\left({a}+\mathrm{1}\right)}}\int\frac{{dp}}{\mathrm{2}\left({p}−\mathrm{1}\right)\sqrt{\frac{{p}}{{a}+\mathrm{1}}−\mathrm{1}}} \\ $$$$\frac{−\mathrm{1}}{\sqrt{\mathrm{8}\left({a}+\mathrm{1}\right)}\:}\int\frac{{dp}}{\left({p}−\mathrm{1}\right)\sqrt{\frac{{p}}{{a}+\mathrm{1}}−\mathrm{1}}} \\ $$$${y}^{\mathrm{2}} =\frac{{p}}{{a}+\mathrm{1}}−\mathrm{1}\:\:\:\:\mathrm{2}{ydy}=\frac{{dp}}{{a}+\mathrm{1}} \\ $$$$=\frac{−\mathrm{1}}{\sqrt{\mathrm{8}\left({a}+\mathrm{1}\right)}}\int\frac{\mathrm{2}\left({a}+\mathrm{1}\right){ydy}}{\left\{\left({a}+\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{1}\right\}×{y}} \\ $$$$=\frac{−\mathrm{2}\left({a}+\mathrm{1}\right)}{\sqrt{\mathrm{8}\left({a}+\mathrm{1}\right)}}\int\frac{{dy}}{\left\{\left({a}+\mathrm{1}\right){y}^{\mathrm{2}} +{a}\right\}} \\ $$$$=\frac{−\left(\sqrt{{a}+\mathrm{1}}\:\right)}{\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{{a}+\mathrm{1}}\int\frac{{dy}}{{y}^{\mathrm{2}} +\frac{{a}}{{a}+\mathrm{1}}} \\ $$$$=\frac{−\mathrm{1}}{\sqrt{\mathrm{2}\left({a}+\mathrm{1}\right)}}×\frac{\mathrm{1}}{\sqrt{\frac{{a}}{{a}+\mathrm{1}}}}{tan}^{−\mathrm{1}} \left(\frac{{y}}{\sqrt{\frac{{a}}{{a}+\mathrm{1}}}}\right)+{c} \\ $$$$\frac{−\mathrm{1}}{\sqrt{\mathrm{2}{a}}}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\frac{{p}}{{a}+\mathrm{1}}−\mathrm{1}}\:}{\sqrt{\frac{{a}}{{a}+\mathrm{1}}}}\right)+{c} \\ $$$$\frac{−\mathrm{1}}{\sqrt{\mathrm{2}{a}}}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\frac{{k}^{\mathrm{2}} }{{a}+\mathrm{1}}−\mathrm{1}}}{\sqrt{\frac{{a}}{{a}+\mathrm{1}}}}\right) \\ $$$$=\frac{−\mathrm{1}}{\sqrt{\mathrm{2}{a}}}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} \left({a}+\mathrm{1}\right)}−\mathrm{1}}\:}{\sqrt{\frac{{a}}{{a}+\mathrm{1}}}}\right) \\ $$$$=\frac{−\mathrm{1}}{\sqrt{\mathrm{2}{a}}}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\left({a}+\mathrm{1}\right)}−\mathrm{1}}}{\sqrt{\frac{{a}}{{a}+\mathrm{1}}}}\right)+{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by MJS last updated on 11/Dec/18

correct, thank you!

$$\mathrm{correct},\:\mathrm{thank}\:\mathrm{you}! \\ $$

Commented by malwaan last updated on 12/Dec/18

great  thank you

$$\mathrm{great} \\ $$$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18

MJS sir your method is short and tricky,  thank you sir...

$${MJS}\:{sir}\:{your}\:{method}\:{is}\:{short}\:{and}\:{tricky}, \\ $$$${thank}\:{you}\:{sir}... \\ $$

Commented by MJS last updated on 12/Dec/18

you′re welcome. I′m an integral addicted...

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}.\:\mathrm{I}'\mathrm{m}\:\mathrm{an}\:\mathrm{integral}\:\mathrm{addicted}... \\ $$

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