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Question Number 109022 by mathdave last updated on 20/Aug/20

	Answered by Dwaipayan Shikari last updated on 20/Aug/20

	$\int_{0}^{\infty} \frac{x + 1}{{(1 + x)}^{3}} - \frac{1}{{(1 + x)}^{3}} d x$ $- {[\frac{1}{1 + x}]}_{0}^{\infty} + {[\frac{1}{2 {(1 + x)}^{2}}]}_{0}^{\infty} = 1 - \frac{1}{2} = \frac{1}{2}$
	Answered by Dwaipayan Shikari last updated on 20/Aug/20

	$4) \int_{0}^{2} \frac{1}{1 + x^{2}} + \frac{x^{4} + x^{2}}{1 + x^{2}} - \frac{x^{2}}{1 + x^{2}}$ ${[{t a n}^{- 1} x + \frac{x^{3}}{3} - x + {t a n}^{- 1} x]}_{0}^{2} = \frac{2}{3} + 2 {t a n}^{- 1} 2$
	Answered by mathmax by abdo last updated on 20/Aug/20
	$2) A =∫_0 ^∞ ((x^2 lnx)/((x^2 +1)^3 )) dx let Q(x) =(x^2 /((x^2 +1)^3 )) ⇒ A =∫_0 ^∞ Q(x)lnxdx =−(1/2)Re(Σ_i Res(Q(z)ln^2 z ,a_i ) let w(z) =Q(z)ln^2 z =((z^2 ln^2 z)/((z^2 +1)^3 )) pole s of w! w(z) =((z^2 ln^2 z)/((z−i)^3 (z+i)^3 )) Res(w,i)=lim_(z→i) (1/((3−1)!)){(z−i)^3 w(z)}^((2)) =(1/2)lim_(z→i) {((z^2 ln^2 z)/((z+i)^3 ))}^((2)) =(1/2)lim_(z→i) { (((2zln^2 z +2((lnz)/z)z^2 )(z+i)^3 −3(z+i)^2 z^2 ln^2 z)/((z+i)^6 ))}^((1)) =(1/2)lim_(z→i) { (((2zln^2 z+2zlnz)(z+i)−3z^2 ln^2 z)/((z+i)^4 ))}^((1)) =(1/2)lim_(z→i) ((2z^2 ln^2 z+2izln^2 z +2z^2 lnz +2iz lnz−3z^2 ln^2 z)/((z+i)^4 )) =(1/2)lim_(z→i) ((zlnz{2ilnz+2i+z))/((z+i)^4 )) =(1/2)lim_(z→i) (((lnz +1)(2ilnz +2i+z)(z+i)^4 −4(z+i)^3 zlnz(2ilnz+2i+z))/((z+i)^8 )) =(1/2)lim_(z→i) (((lnz+1)(2ilnz+2i+z)(z+i)−4zlnz(2ilnz+2i+z))/((z+i)^5 )) =(1/2)(((lni +1)(2ilni+3i)(2i)−4ilni(2ilni +3i))/((2i)^5 )) =(1/2)(((1+((iπ)/2))(2i.((iπ)/2)+3i)(2i)−4i(((iπ)/2))(2i(((iπ)/2)+3i)))/((2i)^5 )) =(1/2)(((1+((iπ)/2))(−π +3i)(2i)+2π(−π +3i))/(32i)) =(1/(64i)){ (1+((iπ)/2))(−2iπ−6)−2π^2 +6iπ} =(1/(64i)){−2iπ−6 +π^2 −2π^2 +6iπ} =(1/(64i)){4iπ−π^2 −6} =(π/(16)) +(1/(64))(π^2 +6)i ....rest calculus of Res(w,−i)...be continued...$
	$2) A = \int_{0}^{\infty} \frac{x^{2} lnx}{{(x^{2} + 1)}^{3}} dx let Q (x) = \frac{x^{2}}{{(x^{2} + 1)}^{3}} \Rightarrow$ $A = \int_{0}^{\infty} Q (x) lnxdx = - \frac{1}{2} Re (\sum_{i} Res (Q (z) \ln^{2} z, a_{i})$ $let w (z) = Q (z) \ln^{2} z = \frac{z^{2} \ln^{2} z}{{(z^{2} + 1)}^{3}} pole s of w!$ $w (z) = \frac{z^{2} \ln^{2} z}{{(z - i)}^{3} {(z + i)}^{3}}$ $Res (w, i) = \lim_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} w (z)}^{(2)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{z^{2} \ln^{2} z}{{(z + i)}^{3}}}^{(2)} = \frac{1}{2} \lim_{z \to i} {\frac{({2 zln}^{2} z + 2 \frac{lnz}{z} z^{2}) {(z + i)}^{3} - 3 {(z + i)}^{2} z^{2} \ln^{2} z}{{(z + i)}^{6}}}^{(1)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{({2 zln}^{2} z + 2 zlnz) (z + i) - {3 z}^{2} \ln^{2} z}{{(z + i)}^{4}}}^{(1)}$ $= \frac{1}{2} \lim_{z \to i} \frac{{2 z}^{2} \ln^{2} z + {2 izln}^{2} z + {2 z}^{2} lnz + 2 iz lnz - {3 z}^{2} \ln^{2} z}{{(z + i)}^{4}}$ $= \frac{1}{2} \lim_{z \to i} \frac{zlnz {2 ilnz + 2 i + z)}{{(z + i)}^{4}}$ $= \frac{1}{2} \lim_{z \to i} \frac{(lnz + 1) (2 ilnz + 2 i + z) {(z + i)}^{4} - 4 {(z + i)}^{3} zlnz (2 ilnz + 2 i + z)}{{(z + i)}^{8}}$ $= \frac{1}{2} \lim_{z \to i} \frac{(lnz + 1) (2 ilnz + 2 i + z) (z + i) - 4 zlnz (2 ilnz + 2 i + z)}{{(z + i)}^{5}}$ $= \frac{1}{2} \frac{(lni + 1) (2 ilni + 3 i) (2 i) - 4 ilni (2 ilni + 3 i)}{{(2 i)}^{5}}$ $= \frac{1}{2} \frac{(1 + \frac{i π}{2}) (2 i . \frac{i π}{2} + 3 i) (2 i) - 4 i (\frac{i π}{2}) (2 i (\frac{i π}{2} + 3 i))}{{(2 i)}^{5}}$ $= \frac{1}{2} \frac{(1 + \frac{i π}{2}) (- π + 3 i) (2 i) + 2 π (- π + 3 i)}{32 i}$ $= \frac{1}{64 i} {(1 + \frac{i π}{2}) (- 2 i π - 6) - 2 π^{2} + 6 i π}$ $= \frac{1}{64 i} {- 2 i π - 6 + π^{2} - 2 π^{2} + 6 i π} = \frac{1}{64 i} {4 i π - π^{2} - 6}$ $= \frac{π}{16} + \frac{1}{64} (π^{2} + 6) i . . . . rest calculus of Res (w, - i) . . . be continued . . .$
	Answered by mathmax by abdo last updated on 20/Aug/20

	$3) I = \int_{0}^{\infty} \frac{x}{{(1 + x)}^{3}} dx changement 1 + x = t give$ $I = \int_{1}^{\infty} \frac{t - 1}{t^{3}} dt = \int_{1}^{\infty} (\frac{1}{t^{2}} - \frac{1}{t^{3}}) dt = {[- \frac{1}{t} + \frac{1}{{2 t}^{2}}]}_{1}^{\infty} = 1 - \frac{1}{2} = \frac{1}{2}$
	Answered by mathmax by abdo last updated on 20/Aug/20

	$4) I = \int_{0}^{2} \frac{1 + x^{4}}{1 + x^{2}} dx \Rightarrow I = \int_{0}^{2} \frac{x^{2} (x^{2} + 1) - x^{2} + 1}{x^{2} + 1} dx$ $= \int_{0}^{2} x^{2} dx - \int_{0}^{2} \frac{x^{2} - 1}{x^{2} + 1} dx = {[\frac{x^{3}}{3}]}_{0}^{2} - \int_{0}^{2} \frac{x^{2} + 1 - 2}{x^{2} + 1} dx$ $= \frac{8}{3} - 2 + 2 {[arctanx]}_{0}^{2} = \frac{2}{3} + 2 \arctan (2)$
	Answered by mathmax by abdo last updated on 20/Aug/20

	$6) I = \int_{0}^{1} \frac{xdx}{1 - x^{2}} \Rightarrow I = \frac{1}{2} \int_{0}^{1} (\frac{1}{1 - x} - \frac{1}{1 + x}) dx$ $= \lim_{ξ \to 1} \frac{1}{2} {[\ln ∣ \frac{1 - x}{1 + x} ∣]}_{0}^{ξ} = + \infty this integral is divergent . . .!$ $\int \sqrt{tanx} dx is solved see the platform$
	Answered by Dwaipayan Shikari last updated on 20/Aug/20

	$- \frac{1}{2} \int_{0}^{1} \frac{- 2 x}{(1 - x^{2})} d x = - \frac{1}{2} {[l o g (1 - x^{2})]}_{0}^{1} \to \infty$
	Answered by mathmax by abdo last updated on 20/Aug/20
	$1) A =∫_0 ^1 ln(((4+3sinx)/(4+3cosx)))dx ⇒A =∫_0 ^1 ln(4+3sinx)dx −∫_0 ^1 ln(4+3cosx)dx =2ln(2)+∫_0 ^1 ln(1+(3/4)sinx)dx−2ln(2) −∫_0 ^1 ln(1+(3/4)cosx)dx =∫_0 ^1 ln(1+(3/4)sinx)−∫_0 ^1 ln(1+(3/4)cosx) let f(a) =∫_0 ^1 ln(1+asinx) with0<a<1 ⇒ f^′ (a) =∫_0 ^1 ((sinx)/(1+asinx))dx =(1/a)∫_0 ^1 ((1+asinx−1)/(1+asinx))dx =(1/a)−(1/a)∫_0 ^1 (dx/(1+asinx)) we have ∫_0 ^1 (dx/(1+asinx)) =_(tan((x/2))=t) ∫_0 ^(tan((1/2))) ((2dt)/((1+t^2 )(1+a((2t)/(1+t^2 ))))) ∫_0 ^(tan((1/2))) ((2dt)/(1+t^2 +2at)) =2 ∫_0 ^(tan((1/2))) (dt/(t^2 +2at +1)) =2∫_0 ^(tan((1/2))) (dt/(t^2 +2at +a^2 +1−a^2 )) =2∫_0 ^(tan((1/2))) (dt/((t+a)^(2 ) +1−a^2 )) =_(t+a =(√(1−a^2 ))u) 2 ∫_(a/(√(1−a^2 ))) ^((tan((1/2))+a)/(√(1−a^2 ))) (((√(1−a^2 ))du)/((1−a^2 )(1+u^2 ))) =(2/(√(1−a^2 ))) [arctanu]_(a/(√(1−a^2 ))) ^((a+tan((1/2)))/(√(1−a^2 ))) =(2/(√(1−a^2 ))){ arctan(((a+tan((1/2)))/(√(1−a^2 )))) −arctan((a/(√(1−a^2 ))))} ⇒ f^′ (a) =(1/a)−(2/(a(√(1−a^2 ))))(arctan(((a+tan((1/2)))/(√(1−a^2 ))))−arctan((a/(√(1−a^2 ))))) ⇒ f(a) =lna−2 ∫_1 ^a (1/(t(√(1−t^2 ))))(arctan(((t+tg((1/2)))/(√(1−t^2 ))))−arctan((t/(√(1−t^2 )))))dt +C c =f(1) ....be continued....$
	$1) A = \int_{0}^{1} \ln (\frac{4 + 3 sinx}{4 + 3 cosx}) dx \Rightarrow A = \int_{0}^{1} \ln (4 + 3 sinx) dx$ $- \int_{0}^{1} \ln (4 + 3 cosx) dx = 2 \ln (2) + \int_{0}^{1} \ln (1 + \frac{3}{4} sinx) dx - 2 \ln (2)$ $- \int_{0}^{1} \ln (1 + \frac{3}{4} cosx) dx = \int_{0}^{1} \ln (1 + \frac{3}{4} sinx) - \int_{0}^{1} \ln (1 + \frac{3}{4} cosx)$ $let f (a) = \int_{0}^{1} \ln (1 + asinx) with 0 < a < 1 \Rightarrow$ $f^{^{'}} (a) = \int_{0}^{1} \frac{sinx}{1 + asinx} dx = \frac{1}{a} \int_{0}^{1} \frac{1 + asinx - 1}{1 + asinx} dx$ $= \frac{1}{a} - \frac{1}{a} \int_{0}^{1} \frac{dx}{1 + asinx} we have \int_{0}^{1} \frac{dx}{1 + asinx} =_{\tan (\frac{x}{2}) = t} \int_{0}^{\tan (\frac{1}{2})} \frac{2 dt}{(1 + t^{2}) (1 + a \frac{2 t}{1 + t^{2}})}$ $\int_{0}^{\tan (\frac{1}{2})} \frac{2 dt}{1 + t^{2} + 2 at} = 2 \int_{0}^{\tan (\frac{1}{2})} \frac{dt}{t^{2} + 2 at + 1} = 2 \int_{0}^{\tan (\frac{1}{2})} \frac{dt}{t^{2} + 2 at + a^{2} + 1 - a^{2}}$ $= 2 \int_{0}^{\tan (\frac{1}{2})} \frac{dt}{{(t + a)}^{2} + 1 - a^{2}} =_{t + a = \sqrt{1 - a^{2}} u} 2 \int_{\frac{a}{\sqrt{1 - a^{2}}}}^{\frac{\tan (\frac{1}{2}) + a}{\sqrt{1 - a^{2}}}} \frac{\sqrt{1 - a^{2}} du}{(1 - a^{2}) (1 + u^{2})}$ $= \frac{2}{\sqrt{1 - a^{2}}} {[arctanu]}_{\frac{a}{\sqrt{1 - a^{2}}}}^{\frac{a + \tan (\frac{1}{2})}{\sqrt{1 - a^{2}}}} = \frac{2}{\sqrt{1 - a^{2}}} {\arctan (\frac{a + \tan (\frac{1}{2})}{\sqrt{1 - a^{2}}})$ $- \arctan (\frac{a}{\sqrt{1 - a^{2}}})} \Rightarrow$ $f^{^{'}} (a) = \frac{1}{a} - \frac{2}{a \sqrt{1 - a^{2}}} (\arctan (\frac{a + \tan (\frac{1}{2})}{\sqrt{1 - a^{2}}}) - \arctan (\frac{a}{\sqrt{1 - a^{2}}})) \Rightarrow$ $f (a) = lna - 2 \int_{1}^{a} \frac{1}{t \sqrt{1 - t^{2}}} (\arctan (\frac{t + tg (\frac{1}{2})}{\sqrt{1 - t^{2}}}) - \arctan (\frac{t}{\sqrt{1 - t^{2}}})) dt + C$ $c = f (1) . . . . be continued . . . .$
	Commented by mathmax by abdo last updated on 20/Aug/20

	$let determine f (a) = \int_{0}^{1} \ln (1 + asinx) dx at form of serie (o < a < 1$ $we have \frac{d}{du} \ln (1 + u) = \frac{1}{1 + u} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} for ∣ u ∣< 1 \Rightarrow$ $\ln (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} u^{n + 1}}{n + 1} + c (c = 0) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} u^{n}}{n} \Rightarrow$ $f (a) = \int_{0}^{1} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {(asinx)}^{n} dx$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} a^{n} \int_{0}^{1} \sin^{n} xdx = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} a^{n} w_{n}$ $with w_{n} = \int_{0}^{1} \sin^{n} xdx (wallis integral on [0, 1] . . . .$ $w_{n} canbe calculsted by recurrence . . .$
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