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Question Number 98426 by mathmax by abdo last updated on 13/Jun/20

let f(x) =∫_(π/4) ^(π/3) (dt/(x+tant)) calculate f(x) 2)explicit g(x) =∫_(π/4) ^(π/3) (dt/((x+tant)^2 )) 3) find the value of integrals ∫_(π/4) ^(π/3) (dt/(2+tant)) and ∫_(π/4) ^(π/3) (dt/((2+tant)^2 ))

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{dt}}{\mathrm{x}+\mathrm{tant}}\:\:\mathrm{calculate}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{explicit}\:\mathrm{g}\left(\mathrm{x}\right)\:=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{dt}}{\left(\mathrm{x}+\mathrm{tant}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{integrals}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{dt}}{\mathrm{2}+\mathrm{tant}}\:\mathrm{and}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{dt}}{\left(\mathrm{2}+\mathrm{tant}\right)^{\mathrm{2}} } \\ $$

Answered by maths mind last updated on 14/Jun/20

f(x)=∫_1 ^(√3) (dy/((x+y)(1+y^2 ))) =(1/(1+x^2 ))∫_1 ^(√3) (((x+y)(x−y)+1+y^2 )/((x+y)(1+y^2 )))dy =(1/(1+x^2 ))∫_1 ^(√3) {(x/(1+y^2 ))−(y/(1+y^2 ))+(1/(x+y))}dy =(x/(1+x^2 ))[(π/(12))]−((ln(2))/(2(1+x^2 )))+((ln(((x+(√3))/(x+1))))/((1+x^2 ))) g(x)=−f′(x) =−{((1−x^2 )/((1+x^2 )^2 ))(π/(12))+((xln(2))/((1+x^2 )^2 ))+(1/((1+x^2 )))[(1/(x+(√3)))−(1/(x+1))]−((2xln(((x+(√3))/(x+1))))/((1+x^2 )^2 ))} 3 ∫(dx/(2+tg(x)))=f(2)=(π/(30))−((ln(2))/(10))+((ln(((2+(√3))/3)))/5) ∫(dx/((2+tg(x))^2 ))=−{((−π)/(100)).+((2ln(2))/(25))+(1/5)((1/(2+(√3)))−(1/3))−((4ln(((2+(√3))/3)))/(25))}

$${f}\left({x}\right)=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{{dy}}{\left({x}+{y}\right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{\left({x}+{y}\right)\left({x}−{y}\right)+\mathrm{1}+{y}^{\mathrm{2}} }{\left({x}+{y}\right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \left\{\frac{{x}}{\mathrm{1}+{y}^{\mathrm{2}} }−\frac{{y}}{\mathrm{1}+{y}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}+{y}}\right\}{dy} \\ $$$$=\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\left[\frac{\pi}{\mathrm{12}}\right]−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}+\frac{{ln}\left(\frac{{x}+\sqrt{\mathrm{3}}}{{x}+\mathrm{1}}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$${g}\left({x}\right)=−{f}'\left({x}\right) \\ $$$$=−\left\{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\frac{\pi}{\mathrm{12}}+\frac{{xln}\left(\mathrm{2}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\left[\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right]−\frac{\mathrm{2}{xln}\left(\frac{{x}+\sqrt{\mathrm{3}}}{{x}+\mathrm{1}}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\right\} \\ $$$$\mathrm{3}\:\:\int\frac{{dx}}{\mathrm{2}+{tg}\left({x}\right)}={f}\left(\mathrm{2}\right)=\frac{\pi}{\mathrm{30}}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{10}}+\frac{{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{3}}\right)}{\mathrm{5}} \\ $$$$\int\frac{{dx}}{\left(\mathrm{2}+{tg}\left({x}\right)\right)^{\mathrm{2}} }=−\left\{\frac{−\pi}{\mathrm{100}}.+\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{\mathrm{25}}+\frac{\mathrm{1}}{\mathrm{5}}\left(\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{4}{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{3}}\right)}{\mathrm{25}}\right\} \\ $$

Commented by abdomathmax last updated on 14/Jun/20

thanks sir.

$$\mathrm{thanks}\:\mathrm{sir}. \\ $$

Commented by maths mind last updated on 14/Jun/20

withe pleaser

$${withe}\:{pleaser}\: \\ $$

Answered by abdomathmax last updated on 14/Jun/20

1) f(x) =∫_(π/4) ^(π/3) (dt/(x+tant)) changement tant =u give f(x) =∫_1 ^(√3) (du/((1+u^2 )(x+u))) let decompose F(u) =(1/((u+x)(u^2 +1))) ⇒F(u) =(a/(u+x)) +((bu +c)/(u^2 +1)) a =(1/(x^2 +1)) lim_(u→+∞) uF(u) =0 =a+b ⇒b=−(1/(x^2 +1)) F(0) =(1/x) =(a/x) +c ⇒1 =a+xc ⇒xc =1−a =1−(1/(x^2 +1)) =(x^2 /(1+x^2 )) ⇒c =(x/(1+x^2 )) ⇒ F(u) =(1/((x^2 +1)(u+x))) +((−(1/(x^2 +1))u +(x/(1+x^2 )))/(u^2 +1)) ⇒f(x) =(1/(x^2 +1)) ∫_1 ^(√3) (du/(u+x)) −(1/(x^2 +1)) ∫_1 ^(√3) ((u−x)/(u^2 +1)) du =(1/(x^2 +1))[ln∣u+x∣]_1 ^(√3) −(1/(2(x^2 +1))) [ln(u^2 +1)]_1 ^(√3) +(x/(x^2 +1)) [arctanu]_1 ^(√3) =(1/(x^2 +1)){ln∣(√3)+x∣−ln∣1+x∣} −((2ln(2))/(2(x^2 +1))) +(x/(x^2 +1))×(π/(12)) =(1/(x^2 +1))ln∣((x+(√3))/(x+1))∣−((ln2)/(x^2 +1)) +((πx)/(12(x^2 +1)))

$$\left.\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{x}\right)\:=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{dt}}{\mathrm{x}+\mathrm{tant}}\:\:\mathrm{changement}\:\mathrm{tant}\:=\mathrm{u}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:\frac{\mathrm{du}}{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\left(\mathrm{x}+\mathrm{u}\right)}\:\mathrm{let}\:\mathrm{decompose} \\ $$$$\mathrm{F}\left(\mathrm{u}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{u}+\mathrm{x}\right)\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow\mathrm{F}\left(\mathrm{u}\right)\:=\frac{\mathrm{a}}{\mathrm{u}+\mathrm{x}}\:+\frac{\mathrm{bu}\:+\mathrm{c}}{\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\mathrm{a}\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:\:\mathrm{lim}_{\mathrm{u}\rightarrow+\infty} \mathrm{uF}\left(\mathrm{u}\right)\:=\mathrm{0}\:=\mathrm{a}+\mathrm{b}\:\Rightarrow\mathrm{b}=−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\mathrm{F}\left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{x}}\:=\frac{\mathrm{a}}{\mathrm{x}}\:+\mathrm{c}\:\Rightarrow\mathrm{1}\:=\mathrm{a}+\mathrm{xc}\:\Rightarrow\mathrm{xc}\:=\mathrm{1}−\mathrm{a} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{c}\:=\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{u}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{u}+\mathrm{x}\right)}\:+\frac{−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{u}\:+\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\frac{\mathrm{du}}{\mathrm{u}+\mathrm{x}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\frac{\mathrm{u}−\mathrm{x}}{\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}}\:\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\left[\mathrm{ln}\mid\mathrm{u}+\mathrm{x}\mid\right]_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\left[\mathrm{ln}\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \\ $$$$+\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:\left[\mathrm{arctanu}\right]_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\left\{\mathrm{ln}\mid\sqrt{\mathrm{3}}+\mathrm{x}\mid−\mathrm{ln}\mid\mathrm{1}+\mathrm{x}\mid\right\} \\ $$$$−\frac{\mathrm{2ln}\left(\mathrm{2}\right)}{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:+\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}×\frac{\pi}{\mathrm{12}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{ln}\mid\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\mathrm{x}+\mathrm{1}}\mid−\frac{\mathrm{ln2}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{\pi\mathrm{x}}{\mathrm{12}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$

Commented by abdomathmax last updated on 14/Jun/20

2) we have f^′ (x) =−∫_(π/4) ^(π/3) (dt/((3+tant)^2 )) =−g(x) ⇒g(x) =−f^′ (x) we have f(x) =(1/(x^2 +1)){ πx−ln2 +ln∣((x+(√3))/(x+1))∣} ⇒ f^′ (x) =−((2x)/((x^(2 ) +1)^2 )){ πx−ln(2)+ln∣((x+(√3))/(x+1))∣} +(1/(x^2 +1)){ π+(((((x+(√3))/(x+1)))^′ )/((x+(√3))/(x+1)))} =((−2x)/((x^2 +1)^2 )){πx−ln2 +ln∣((x+(√3))/(x+1))∣} +(1/(x^2 +1)){ π +(((x+1))/(x+(√3)))×((x+1−(x+(√3)))/((x+1)^2 ))} =((−2x)/((x^2 +1)^2 )){πx−ln2 +ln∣((x+(√3))/(x+1))∣} +(1/(x^2 +1)){ π +((1−(√3))/((x+(√3))(x+1)))}

$$\left.\mathrm{2}\right)\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}^{'} \left(\mathrm{x}\right)\:=−\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\frac{\mathrm{dt}}{\left(\mathrm{3}+\mathrm{tant}\right)^{\mathrm{2}} }\:=−\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\Rightarrow\mathrm{g}\left(\mathrm{x}\right)\:=−\mathrm{f}^{'} \left(\mathrm{x}\right)\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\left\{\:\pi\mathrm{x}−\mathrm{ln2}\:+\mathrm{ln}\mid\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\mathrm{x}+\mathrm{1}}\mid\right\}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)\:=−\frac{\mathrm{2x}}{\left(\mathrm{x}^{\mathrm{2}\:} +\mathrm{1}\right)^{\mathrm{2}} }\left\{\:\pi\mathrm{x}−\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{ln}\mid\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\mathrm{x}+\mathrm{1}}\mid\right\} \\ $$$$+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\left\{\:\pi+\frac{\left(\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\mathrm{x}+\mathrm{1}}\right)^{'} }{\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\mathrm{x}+\mathrm{1}}}\right\} \\ $$$$=\frac{−\mathrm{2x}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\left\{\pi\mathrm{x}−\mathrm{ln2}\:+\mathrm{ln}\mid\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\mathrm{x}+\mathrm{1}}\mid\right\} \\ $$$$+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\left\{\:\pi\:+\frac{\left(\mathrm{x}+\mathrm{1}\right)}{\mathrm{x}+\sqrt{\mathrm{3}}}×\frac{\mathrm{x}+\mathrm{1}−\left(\mathrm{x}+\sqrt{\mathrm{3}}\right)}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{−\mathrm{2x}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\left\{\pi\mathrm{x}−\mathrm{ln2}\:+\mathrm{ln}\mid\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\mathrm{x}+\mathrm{1}}\mid\right\} \\ $$$$+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\left\{\:\pi\:+\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\left(\mathrm{x}+\sqrt{\mathrm{3}}\right)\left(\mathrm{x}+\mathrm{1}\right)}\right\} \\ $$

Commented by mathmax by abdo last updated on 14/Jun/20

⇒g(x) =((2x)/((x^2 +1)^2 )){πx−ln2 +ln∣((x+(√3))/(x+1))∣}−(1/(x^2 +1)){π +((1−(√3))/((x+(√3))(x+1)))}

$$\Rightarrow\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{2x}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\left\{\pi\mathrm{x}−\mathrm{ln2}\:+\mathrm{ln}\mid\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\mathrm{x}+\mathrm{1}}\mid\right\}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\left\{\pi\:+\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\left(\mathrm{x}+\sqrt{\mathrm{3}}\right)\left(\mathrm{x}+\mathrm{1}\right)}\right\} \\ $$

Commented by mathmax by abdo last updated on 14/Jun/20

3) ∫_(π/4) ^(π/3) (dt/(2+tant)) =f(2) =(1/5)ln(((2+(√3))/3))−((ln2)/5) +((2π)/(12(5))) =(1/5)ln(((2+(√3))/3))−((ln(2))/5) +(π/(30))

$$\left.\mathrm{3}\right)\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\frac{\mathrm{dt}}{\mathrm{2}+\mathrm{tant}}\:=\mathrm{f}\left(\mathrm{2}\right)\:=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{3}}\right)−\frac{\mathrm{ln2}}{\mathrm{5}}\:+\frac{\mathrm{2}\pi}{\mathrm{12}\left(\mathrm{5}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{3}}\right)−\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{5}}\:+\frac{\pi}{\mathrm{30}} \\ $$

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