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Question Number 49635 by maxmathsup by imad last updated on 08/Dec/18

1)find f(x) =∫_0 ^(π/4) ((sint)/(2+x cos(2t)))dt 2) find g(x) =∫_0 ^(π/4) ((sint sin(2t)/((2+x cos(2t))^2 ))dx 3) find the value of ∫_0 ^(π/4) ((sint)/(2+3 cos(2t)))dt and ∫_0 ^(π/4) ((sin(t)sin(2t))/((2+3cos(2t))^2 ))dt

$$\left.\mathrm{1}\right){find}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sint}}{\mathrm{2}+{x}\:{cos}\left(\mathrm{2}{t}\right)}{dt} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sint}\:{sin}\left(\mathrm{2}{t}\right.}{\left(\mathrm{2}+{x}\:{cos}\left(\mathrm{2}{t}\right)\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{sint}}{\mathrm{2}+\mathrm{3}\:{cos}\left(\mathrm{2}{t}\right)}{dt}\:\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{sin}\left({t}\right){sin}\left(\mathrm{2}{t}\right)}{\left(\mathrm{2}+\mathrm{3}{cos}\left(\mathrm{2}{t}\right)\right)^{\mathrm{2}} }{dt} \\ $$

Commented by Abdo msup. last updated on 15/Dec/18

1) we have f(x) =∫_0 ^(π/4) ((sint)/(2+x(2cos^2 t−1)))dt =∫_0 ^(π/4) ((sint)/(2xcos^2 t +2−x))dt =_(cost =u) ∫_1 ^(1/(√2)) ((−du)/(2xu^2 +2−x)) =−∫_1 ^(1/(√2)) (du/(2x(u^2 +((2−x)/(2x))))) =(1/(2x)) ∫_((√2)/2) ^1 (du/(u^2 +((2−x)/(2x)))) if ((2−x)/(2x)) >0 we do the changement u=(√((2−x)/(2x)))α f(x)=(1/(2x)) ∫_(((√2)/2)(√((2x)/(2−x)))) ^(√((2x)/(2−x))) (1/(((2−x)/(2x))(1+α^2 ))) (√((2−x)/(2x)))dα =(1/(2−x))(√((2−x)/(2x))) [arctan(α)]_(((√2)/2)(√((2x)/(2−x)))) ^(√((2x)/(2−x))) f(x)= (1/(√((2−x)2x))) { arctan((√((2x)/(2−x))))−arctan(((√2)/2)(√((2x)/(2−x))))} if ((2−x)/(2x))<0 we get f(x)=(1/(2x)) ∫_((√2)/2) ^1 (du/(u^2 −((x−2)/(2x)))) and wedo the changement u=(√((x−2)/(2x)))α and that lead to f(x)=(1/(2x)) ∫_(((√2)/2)(√((x−2)/(2x)))) ^(√((x−2)/(2x))) (1/(((x−2)/(2x))(α^2 −1)))(√((x−2)/(2x)))dα = (1/(√((x−2)2x))) (1/2)∫_(..) ^(..) {(1/(α−1)) −(1/(α+1))}dα =(1/(2(√(2x(x−2)))))[ln∣((α−1)/(α+1))∣]_(((√2)/2)(√((x−2)/(2x)))) ^(√((x−2)/(2x))) =(1/(2(√(2x(x−2))))) {ln∣(((√((x−2)/(2x)))−1)/((√((x−2)/(2x)))+1))∣−ln∣((((√2)/2)(√((x−2)/(2x)))−1)/(((√2)/2)(√((x−2)/(2x)))+1))∣.

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{sint}}{\mathrm{2}+{x}\left(\mathrm{2}{cos}^{\mathrm{2}} {t}−\mathrm{1}\right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{sint}}{\mathrm{2}{xcos}^{\mathrm{2}} {t}\:+\mathrm{2}−{x}}{dt}\:=_{{cost}\:={u}} \:\:\:\:\:\:\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \:\:\:\:\:\:\frac{−{du}}{\mathrm{2}{xu}^{\mathrm{2}} \:+\mathrm{2}−{x}} \\ $$$$=−\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \:\:\:\:\:\:\frac{{du}}{\mathrm{2}{x}\left({u}^{\mathrm{2}} \:+\frac{\mathrm{2}−{x}}{\mathrm{2}{x}}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}{x}}\:\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\frac{\mathrm{2}−{x}}{\mathrm{2}{x}}} \\ $$$${if}\:\frac{\mathrm{2}−{x}}{\mathrm{2}{x}}\:>\mathrm{0}\:\:{we}\:{do}\:{the}\:{changement}\:{u}=\sqrt{\frac{\mathrm{2}−{x}}{\mathrm{2}{x}}}\alpha \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{x}}\:\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\frac{\mathrm{2}{x}}{\mathrm{2}−{x}}}} ^{\sqrt{\frac{\mathrm{2}{x}}{\mathrm{2}−{x}}}} \:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}−{x}}{\mathrm{2}{x}}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\:\sqrt{\frac{\mathrm{2}−{x}}{\mathrm{2}{x}}}{d}\alpha \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}−{x}}\sqrt{\frac{\mathrm{2}−{x}}{\mathrm{2}{x}}}\:\:\left[{arctan}\left(\alpha\right)\right]_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\frac{\mathrm{2}{x}}{\mathrm{2}−{x}}}} ^{\sqrt{\frac{\mathrm{2}{x}}{\mathrm{2}−{x}}}} \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\sqrt{\left(\mathrm{2}−{x}\right)\mathrm{2}{x}}}\:\left\{\:{arctan}\left(\sqrt{\frac{\mathrm{2}{x}}{\mathrm{2}−{x}}}\right)−{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\frac{\mathrm{2}{x}}{\mathrm{2}−{x}}}\right)\right\} \\ $$$${if}\:\frac{\mathrm{2}−{x}}{\mathrm{2}{x}}<\mathrm{0}\:\:{we}\:{get}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{x}}\:\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\:\:\frac{{du}}{{u}^{\mathrm{2}} −\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}}\:{and}\:{wedo} \\ $$$${the}\:{changement}\:{u}=\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}}\alpha\:\:{and}\:{that}\:{lead}\:{to}\: \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{x}}\:\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}}} ^{\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}}} \:\:\:\:\:\:\:\frac{\mathrm{1}}{\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}}{d}\alpha \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\left({x}−\mathrm{2}\right)\mathrm{2}{x}}}\:\frac{\mathrm{1}}{\mathrm{2}}\int_{..} ^{..} \:\:\left\{\frac{\mathrm{1}}{\alpha−\mathrm{1}}\:−\frac{\mathrm{1}}{\alpha+\mathrm{1}}\right\}{d}\alpha \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}{x}\left({x}−\mathrm{2}\right)}}\left[{ln}\mid\frac{\alpha−\mathrm{1}}{\alpha+\mathrm{1}}\mid\right]_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}}} ^{\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}{x}\left({x}−\mathrm{2}\right)}}\:\left\{{ln}\mid\frac{\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}}−\mathrm{1}}{\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}}+\mathrm{1}}\mid−{ln}\mid\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}}−\mathrm{1}}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{2}{x}}}+\mathrm{1}}\mid.\right. \\ $$$$ \\ $$

Answered by Smail last updated on 09/Dec/18

f(x)=∫_0 ^(π/4) ((sint)/(2+xcos(2t)))dt =∫_0 ^(π/4) ((sint)/(2+x(2cos^2 (t)−1)))dt u=cost⇒du=−sintdt f(x)=−∫_1 ^((√2)/2) (du/(2+2xu^2 −x)) =−∫_1 ^(1/(√2)) (du/((2−x)(((2xu^2 )/(2−x))+1))) y=(√((2x)/(2−x)))u⇒du=(√((2−x)/(2x)))dy f(x)=−(1/(2−x))×(√((2−x)/(2x)))∫_(√(2x/(2−x))) ^(√(x/(2−x))) (dy/(y^2 +1)) =−(1/(√(2x(2−x))))[tan^(−1) ((√(x/(2−x))))−tan^(−1) ((√((2x)/(2−x))))]

$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \frac{{sint}}{\mathrm{2}+{xcos}\left(\mathrm{2}{t}\right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \frac{{sint}}{\mathrm{2}+{x}\left(\mathrm{2}{cos}^{\mathrm{2}} \left({t}\right)−\mathrm{1}\right)}{dt} \\ $$$${u}={cost}\Rightarrow{du}=−{sintdt} \\ $$$${f}\left({x}\right)=−\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}/\mathrm{2}} \frac{{du}}{\mathrm{2}+\mathrm{2}{xu}^{\mathrm{2}} −{x}} \\ $$$$=−\int_{\mathrm{1}} ^{\mathrm{1}/\sqrt{\mathrm{2}}} \frac{{du}}{\left(\mathrm{2}−{x}\right)\left(\frac{\mathrm{2}{xu}^{\mathrm{2}} }{\mathrm{2}−{x}}+\mathrm{1}\right)} \\ $$$${y}=\sqrt{\frac{\mathrm{2}{x}}{\mathrm{2}−{x}}}{u}\Rightarrow{du}=\sqrt{\frac{\mathrm{2}−{x}}{\mathrm{2}{x}}}{dy} \\ $$$${f}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}−{x}}×\sqrt{\frac{\mathrm{2}−{x}}{\mathrm{2}{x}}}\int_{\sqrt{\mathrm{2}{x}/\left(\mathrm{2}−{x}\right)}} ^{\sqrt{{x}/\left(\mathrm{2}−{x}\right)}} \frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}{x}\left(\mathrm{2}−{x}\right)}}\left[{tan}^{−\mathrm{1}} \left(\sqrt{\frac{{x}}{\mathrm{2}−{x}}}\right)−{tan}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{2}{x}}{\mathrm{2}−{x}}}\right)\right] \\ $$$$ \\ $$

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