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 ∫


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

$let f (x) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{x + tant} calculate f (x)$ $2) explicit g (x) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{{(x + tant)}^{2}}$ $3) find the value of integrals \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{2 + tant} and \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{{(2 + tant)}^{2}}$
	Answered by maths mind last updated on 14/Jun/20
	$f(x)=∫_1 ^(√3) (dy/((x+y)(1+y^2 ))) =(1/(1+x^2 ))∫_1 ^(√3) (((x+y)(x−y)+1+y^2 )/((x+y)(1+y^2 )))dy =(1/(1+x^2 ))∫_1 ^(√3) {(x/(1+y^2 ))−(y/(1+y^2 ))+(1/(x+y))}dy =(x/(1+x^2 ))[(π/(12))]−((ln(2))/(2(1+x^2 )))+((ln(((x+(√3))/(x+1))))/((1+x^2 ))) g(x)=−f′(x) =−{((1−x^2 )/((1+x^2 )^2 ))(π/(12))+((xln(2))/((1+x^2 )^2 ))+(1/((1+x^2 )))[(1/(x+(√3)))−(1/(x+1))]−((2xln(((x+(√3))/(x+1))))/((1+x^2 )^2 ))} 3 ∫(dx/(2+tg(x)))=f(2)=(π/(30))−((ln(2))/(10))+((ln(((2+(√3))/3)))/5) ∫(dx/((2+tg(x))^2 ))=−{((−π)/(100)).+((2ln(2))/(25))+(1/5)((1/(2+(√3)))−(1/3))−((4ln(((2+(√3))/3)))/(25))}$
	$f (x) = \int_{1}^{\sqrt{3}} \frac{d y}{(x + y) (1 + y^{2})}$ $= \frac{1}{1 + x^{2}} \int_{1}^{\sqrt{3}} \frac{(x + y) (x - y) + 1 + y^{2}}{(x + y) (1 + y^{2})} d y$ $= \frac{1}{1 + x^{2}} \int_{1}^{\sqrt{3}} {\frac{x}{1 + y^{2}} - \frac{y}{1 + y^{2}} + \frac{1}{x + y}} d y$ $= \frac{x}{1 + x^{2}} [\frac{π}{12}] - \frac{l n (2)}{2 (1 + x^{2})} + \frac{l n (\frac{x + \sqrt{3}}{x + 1})}{(1 + x^{2})}$ $g (x) = - f^{'} (x)$ $= - {\frac{1 - x^{2}}{{(1 + x^{2})}^{2}} \frac{π}{12} + \frac{x l n (2)}{{(1 + x^{2})}^{2}} + \frac{1}{(1 + x^{2})} [\frac{1}{x + \sqrt{3}} - \frac{1}{x + 1}] - \frac{2 x l n (\frac{x + \sqrt{3}}{x + 1})}{{(1 + x^{2})}^{2}}}$ $3 \int \frac{d x}{2 + t g (x)} = f (2) = \frac{π}{30} - \frac{l n (2)}{10} + \frac{l n (\frac{2 + \sqrt{3}}{3})}{5}$ $\int \frac{d x}{{(2 + t g (x))}^{2}} = - {\frac{- π}{100} . + \frac{2 l n (2)}{25} + \frac{1}{5} (\frac{1}{2 + \sqrt{3}} - \frac{1}{3}) - \frac{4 l n (\frac{2 + \sqrt{3}}{3})}{25}}$
	Commented by abdomathmax last updated on 14/Jun/20

	$thanks sir .$
	Commented by maths mind last updated on 14/Jun/20

	$w i t h e p l e a s e r$
	Answered by abdomathmax last updated on 14/Jun/20
	$1) f(x) =∫_(π/4) ^(π/3) (dt/(x+tant)) changement tant =u give f(x) =∫_1 ^(√3) (du/((1+u^2 )(x+u))) let decompose F(u) =(1/((u+x)(u^2 +1))) ⇒F(u) =(a/(u+x)) +((bu +c)/(u^2 +1)) a =(1/(x^2 +1)) lim_(u→+∞) uF(u) =0 =a+b ⇒b=−(1/(x^2 +1)) F(0) =(1/x) =(a/x) +c ⇒1 =a+xc ⇒xc =1−a =1−(1/(x^2 +1)) =(x^2 /(1+x^2 )) ⇒c =(x/(1+x^2 )) ⇒ F(u) =(1/((x^2 +1)(u+x))) +((−(1/(x^2 +1))u +(x/(1+x^2 )))/(u^2 +1)) ⇒f(x) =(1/(x^2 +1)) ∫_1 ^(√3) (du/(u+x)) −(1/(x^2 +1)) ∫_1 ^(√3) ((u−x)/(u^2 +1)) du =(1/(x^2 +1))[ln∣u+x∣]_1 ^(√3) −(1/(2(x^2 +1))) [ln(u^2 +1)]_1 ^(√3) +(x/(x^2 +1)) [arctanu]_1 ^(√3) =(1/(x^2 +1)){ln∣(√3)+x∣−ln∣1+x∣} −((2ln(2))/(2(x^2 +1))) +(x/(x^2 +1))×(π/(12)) =(1/(x^2 +1))ln∣((x+(√3))/(x+1))∣−((ln2)/(x^2 +1)) +((πx)/(12(x^2 +1)))$
	$1) f (x) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{x + tant} changement tant = u give$ $f (x) = \int_{1}^{\sqrt{3}} \frac{du}{(1 + u^{2}) (x + u)} let decompose$ $F (u) = \frac{1}{(u + x) (u^{2} + 1)} \Rightarrow F (u) = \frac{a}{u + x} + \frac{bu + c}{u^{2} + 1}$ $a = \frac{1}{x^{2} + 1} \lim_{u \to + \infty} uF (u) = 0 = a + b \Rightarrow b = - \frac{1}{x^{2} + 1}$ $F (0) = \frac{1}{x} = \frac{a}{x} + c \Rightarrow 1 = a + xc \Rightarrow xc = 1 - a$ $= 1 - \frac{1}{x^{2} + 1} = \frac{x^{2}}{1 + x^{2}} \Rightarrow c = \frac{x}{1 + x^{2}} \Rightarrow$ $F (u) = \frac{1}{(x^{2} + 1) (u + x)} + \frac{- \frac{1}{x^{2} + 1} u + \frac{x}{1 + x^{2}}}{u^{2} + 1}$ $\Rightarrow f (x) = \frac{1}{x^{2} + 1} \int_{1}^{\sqrt{3}} \frac{du}{u + x}$ $- \frac{1}{x^{2} + 1} \int_{1}^{\sqrt{3}} \frac{u - x}{u^{2} + 1} du$ $= \frac{1}{x^{2} + 1} {[\ln ∣ u + x ∣]}_{1}^{\sqrt{3}} - \frac{1}{2 (x^{2} + 1)} {[\ln (u^{2} + 1)]}_{1}^{\sqrt{3}}$ $+ \frac{x}{x^{2} + 1} {[arctanu]}_{1}^{\sqrt{3}}$ $= \frac{1}{x^{2} + 1} {\ln ∣ \sqrt{3} + x ∣ - \ln ∣ 1 + x ∣}$ $- \frac{2 \ln (2)}{2 (x^{2} + 1)} + \frac{x}{x^{2} + 1} \times \frac{π}{12}$ $= \frac{1}{x^{2} + 1} \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣ - \frac{\ln 2}{x^{2} + 1} + \frac{π x}{12 (x^{2} + 1)}$
	Commented by abdomathmax last updated on 14/Jun/20
	$2) we have f^′ (x) =−∫_(π/4) ^(π/3) (dt/((3+tant)^2 )) =−g(x) ⇒g(x) =−f^′ (x) we have f(x) =(1/(x^2 +1)){ πx−ln2 +ln∣((x+(√3))/(x+1))∣} ⇒ f^′ (x) =−((2x)/((x^(2 ) +1)^2 )){ πx−ln(2)+ln∣((x+(√3))/(x+1))∣} +(1/(x^2 +1)){ π+(((((x+(√3))/(x+1)))^′ )/((x+(√3))/(x+1)))} =((−2x)/((x^2 +1)^2 )){πx−ln2 +ln∣((x+(√3))/(x+1))∣} +(1/(x^2 +1)){ π +(((x+1))/(x+(√3)))×((x+1−(x+(√3)))/((x+1)^2 ))} =((−2x)/((x^2 +1)^2 )){πx−ln2 +ln∣((x+(√3))/(x+1))∣} +(1/(x^2 +1)){ π +((1−(√3))/((x+(√3))(x+1)))}$
	$2) we have f^{^{'}} (x) = - \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{{(3 + tant)}^{2}} = - g (x)$ $\Rightarrow g (x) = - f^{^{'}} (x) we have$ $f (x) = \frac{1}{x^{2} + 1} {π x - \ln 2 + \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣} \Rightarrow$ $f^{^{'}} (x) = - \frac{2 x}{{(x^{2} + 1)}^{2}} {π x - \ln (2) + \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣}$ $+ \frac{1}{x^{2} + 1} {π + \frac{{(\frac{x + \sqrt{3}}{x + 1})}^{^{'}}}{\frac{x + \sqrt{3}}{x + 1}}}$ $= \frac{- 2 x}{{(x^{2} + 1)}^{2}} {π x - \ln 2 + \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣}$ $+ \frac{1}{x^{2} + 1} {π + \frac{(x + 1)}{x + \sqrt{3}} \times \frac{x + 1 - (x + \sqrt{3})}{{(x + 1)}^{2}}}$ $= \frac{- 2 x}{{(x^{2} + 1)}^{2}} {π x - \ln 2 + \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣}$ $+ \frac{1}{x^{2} + 1} {π + \frac{1 - \sqrt{3}}{(x + \sqrt{3}) (x + 1)}}$
	Commented by mathmax by abdo last updated on 14/Jun/20
	$⇒g(x) =((2x)/((x^2 +1)^2 )){πx−ln2 +ln∣((x+(√3))/(x+1))∣}−(1/(x^2 +1)){π +((1−(√3))/((x+(√3))(x+1)))}$
	$\Rightarrow g (x) = \frac{2 x}{{(x^{2} + 1)}^{2}} {π x - \ln 2 + \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣} - \frac{1}{x^{2} + 1} {π + \frac{1 - \sqrt{3}}{(x + \sqrt{3}) (x + 1)}}$
	Commented by mathmax by abdo last updated on 14/Jun/20

	$3) \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{2 + tant} = f (2) = \frac{1}{5} \ln (\frac{2 + \sqrt{3}}{3}) - \frac{\ln 2}{5} + \frac{2 π}{12 (5)}$ $= \frac{1}{5} \ln (\frac{2 + \sqrt{3}}{3}) - \frac{\ln (2)}{5} + \frac{π}{30}$
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