Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 58487 by Mr X pcx last updated on 23/Apr/19

let f(x) =∫_(π/4) ^(π/3) (dt/(2+xsint)) 1) find a explicit form of f(x) 2)determine also g(x)=∫_(π/4) ^(π/3) ((sint)/((2+xsint)^2 ))dt 3) find the value of ∫_(π/4) ^(π/3) (dt/(2+3sint)) and ∫_(π/4) ^(π/3) ((sint)/((2+3sint)^2 ))dt

$${let}\:{f}\left({x}\right)\:=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\frac{{dt}}{\mathrm{2}+{xsint}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){determine}\:{also}\:{g}\left({x}\right)=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\frac{{sint}}{\left(\mathrm{2}+{xsint}\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\frac{{dt}}{\mathrm{2}+\mathrm{3}{sint}} \\ $$$${and}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\frac{{sint}}{\left(\mathrm{2}+\mathrm{3}{sint}\right)^{\mathrm{2}} }{dt} \\ $$

Commented by maxmathsup by imad last updated on 24/Apr/19

changement tan((t/2)) =u give f(x) =∫_((√2)−1) ^(1/(√3)) ((2du)/((1+u^2 )(2+x ((2u)/(1+u^2 ))))) = ∫_((√2)−1) ^(1/(√3)) ((2du)/(2(1+u^2 ) +2xu)) = ∫_((√2)−1) ^(1/(√3)) (du/(u^2 +2xu +1)) =∫_((√2)−1) ^(1/(√3)) (du/((u+x)^2 +1−x^2 )) case 1 if 1−x^2 >0 ⇒∣x∣<1 we do the changement u+x =(√(1−x^2 ))α ⇒f(x) = ∫_(((√2)−1 +x)/(√(1−x^2 ))) ^(((1/(√3))+x)/(√(1−x^2 ))) (((√(1−x^2 ))dα)/((1−x^2 )(1+α^2 ))) =(1/(√(1−x^2 ))) [ arctanα]_(((√2)−1+x)/(√(1−x^2 ))) ^(((1/(√3))+x)/(√(1−x^2 ))) ⇒f(x) =(1/(√(1−x^2 ))){ arctan(((1+x(√3))/((√3)(√(1−x^2 )))))−arctan((((√2)−1+x)/(√(1−x^2 ))))} case 2 if 1−x^2 <0 ⇒∣x∣>1 ⇒f(x)=∫_((√2)−1) ^(1/(√3)) (du/((u+x)^2 −((√(x^2 −1)))^2 )) =∫_((√2)−1) ^(1/(√3)) (du/((u+x+(√(x^2 −1)))(u+x−(√(x^2 −1))))) =(1/(2(√(x^2 −1))))∫_((√2)−1) ^(1/(√3)) {(1/(u+x −(√(x^2 −1)))) −(1/(u+x +(√(x^2 −1))))}du =(1/(2(√(x^2 −1)))) [ln∣((u+x −(√(x^2 −1)))/(u+x +(√(x^2 −1)))) ∣]_(u=(√2)−1) ^(u=(1/(√3))) =(1/(2(√(x^2 −1)))){ln∣(((1/((√3) ))+x−(√(x^2 −1)))/((1/(√3))+x+(√(x^2 −1))))∣−ln∣(((√2)−1+x −(√(x^2 −1)))/((√2)−1+x +(√(x^2 −1))))∣ } .

$${changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)\:={u}\:\:{give}\:\:{f}\left({x}\right)\:=\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}+{x}\:\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\right)} \\ $$$$=\:\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\frac{\mathrm{2}{du}}{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\:+\mathrm{2}{xu}}\:=\:\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}{xu}\:+\mathrm{1}}\:\:=\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\frac{{du}}{\left({u}+{x}\right)^{\mathrm{2}} \:+\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${case}\:\mathrm{1}\:\:{if}\:\:\mathrm{1}−{x}^{\mathrm{2}} >\mathrm{0}\:\Rightarrow\mid{x}\mid<\mathrm{1}\:\:\:\:{we}\:{do}\:{the}\:{changement}\:{u}+{x}\:=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\alpha \\ $$$$\Rightarrow{f}\left({x}\right)\:=\:\int_{\frac{\sqrt{\mathrm{2}}−\mathrm{1}\:+{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} ^{\frac{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}+{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} \:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{d}\alpha}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\left[\:{arctan}\alpha\right]_{\frac{\sqrt{\mathrm{2}}−\mathrm{1}+{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} ^{\frac{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}+{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} \:\:\Rightarrow{f}\left({x}\right)\:=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left\{\:{arctan}\left(\frac{\mathrm{1}+{x}\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)−{arctan}\left(\frac{\sqrt{\mathrm{2}}−\mathrm{1}+{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)\right\} \\ $$$${case}\:\mathrm{2}\:\:{if}\:\mathrm{1}−{x}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\mid{x}\mid>\mathrm{1}\:\:\Rightarrow{f}\left({x}\right)=\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\frac{{du}}{\left({u}+{x}\right)^{\mathrm{2}} −\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} } \\ $$$$=\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\:\frac{{du}}{\left({u}+{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\left({u}+{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\left\{\frac{\mathrm{1}}{{u}+{x}\:−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:−\frac{\mathrm{1}}{{u}+{x}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\right\}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\left[{ln}\mid\frac{{u}+{x}\:−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{u}+{x}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\mid\right]_{{u}=\sqrt{\mathrm{2}}−\mathrm{1}} ^{{u}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\left\{{ln}\mid\frac{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}\:}+{x}−\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}}{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}+{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\mid−{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}+{x}\:−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\sqrt{\mathrm{2}}−\mathrm{1}+{x}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\mid\:\right\}\:. \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 25/Apr/19

2)we have f^′ (x) =∫_(π/4) ^(π/3) −((sint)/((2+xsint)^2 )) dt =−g(x) ⇒g(x) =−f^′ (x) rest to calculate f^′ (x)

$$\left.\mathrm{2}\right){we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:−\frac{{sint}}{\left(\mathrm{2}+{xsint}\right)^{\mathrm{2}} }\:{dt}\:=−{g}\left({x}\right)\:\Rightarrow{g}\left({x}\right)\:=−{f}^{'} \left({x}\right)\:\:{rest}\:{to}\:{calculate} \\ $$$${f}^{'} \left({x}\right) \\ $$

Commented by maxmathsup by imad last updated on 25/Apr/19

3) ∫_(π/4) ^(π/3) (dt/(2+3cost)) =f(3) =(1/(2(√(3^2 −1)))){ln∣(((1/(√3))+3−(√(3^2 −1)))/((1/(√3)) +3 +(√(3^2 −1))))∣−ln∣(((√2)−1 +3−(√(3^2 −1)))/((√2)−1 +3+(√(3^2 −1))))∣ =(1/(4(√2))){ln∣(((1/(√3))+3−2(√2))/((1/((√3) )) +3+2(√2)))∣ −ln∣((2−(√2))/(2+3(√2)))∣ .

$$\left.\mathrm{3}\right)\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\frac{{dt}}{\mathrm{2}+\mathrm{3}{cost}}\:={f}\left(\mathrm{3}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}}\left\{{ln}\mid\frac{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}+\mathrm{3}−\sqrt{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}}{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:+\mathrm{3}\:+\sqrt{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}}\mid−{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}\:+\mathrm{3}−\sqrt{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}}{\sqrt{\mathrm{2}}−\mathrm{1}\:+\mathrm{3}+\sqrt{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}}\mid\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\left\{{ln}\mid\frac{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}\:}\:+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\mid\:−{ln}\mid\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}+\mathrm{3}\sqrt{\mathrm{2}}}\mid\:.\right. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com