Question Number 108145 by bemath last updated on 15/Aug/20
$$\:\:\:\frac{\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc}{\Pi} \\ $$ $$\:\left(\mathrm{1}\right)\:\:\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{4}}\right)\:>\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{4}}\right)\: \\ $$ $$\:\left(\mathrm{2}\right)\:\int\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx} \\ $$ $$\left(\mathrm{3}\right)\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\sqrt{\mathrm{tan}\:{x}}}{\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:{dx} \\ $$
Commented bybemath last updated on 15/Aug/20
dear admint tinkutara. i have used v.2.138. problem for which i can not edit answer
Commented byTinku Tara last updated on 15/Aug/20
$$\mathrm{Just}\:\mathrm{tested}.\:\mathrm{What}\:\mathrm{error}\:\mathrm{message} \\ $$ $$\mathrm{do}\:\mathrm{u}\:\mathrm{get}? \\ $$
Answered by 1549442205PVT last updated on 15/Aug/20
![1)cos^2 (x/4)>((√2)/2)+sin^2 (x/4)⇔cos^2 (x/4)−sin^2 (x/4)>((√2)/2) ⇔cos(x/2)>((√2)/2)=cos(π/4) ⇔2k𝛑−(𝛑/4)<x<(𝛑/4)+2k𝛑 2)Put x=cosθ⇒dx=−sinθdθ I=−∫(√((1+cosθ)/(1−cosθ)))sinθ=−∫(√((2cos^2 (θ/2))/(2sin^2 (θ/2))))×((2tan(θ/2))/(1+tan^2 (θ/2)))dθ =−∫cot(θ/2)×((2tan(θ/2))/(1+tan^2 (θ/2)))dθ=−2∫(dθ/(1+tan^2 (θ/2))) =−2∫cos^2 (θ/2)dθ=−∫(1+cosθ)dθ =−𝛉−sin𝛉=−sin^(−1) x−(√(1−x^2 )) +C 3)∫_0 ^(π/2) ((√(tanx))/((cosx+sinx)^2 ))dx=∫((√(tanx))/(1+2sin2x))dx =∫_0 ^(π/2) ((√(tanx))/(1+((2tanx)/(1+tan^2 x))))dx.Put tanx=t ⇒dt=(1+t^2 )dx,((√(tanx))/(1+((2tanx)/(1+tan^2 x))))=(((1+t^2 )(√t))/((1+t)^2 )) ⇒I=∫_0 ^∞ (((1+t^2 )^2 (√t))/((1+t)^2 ))dt=∫_0 ^∞ (([(1+t)^2 −2t]^2 (√t))/((1+t)^2 )) =∫_0 ^∞ [(1+t)^2 −4t+((4t^2 )/((1+t)^2 ))](√t) dt =∫_0 ^∞ [(1−2t+t^2 )(√t) +((4t^2 (√t))/((1+t)^2 ))]dt =((2/3)t^(3/2) −(4/5)t^(5/2) +(2/7)t^(7/2) )+∫_0 ^∞ ((4t^2 (√t))/((1+t)^2 ))dt Put (√t)=u⇒du=(1/2(√t))dt⇒dt=2udu A=∫_0 ^∞ ((4t^2 (√t))/((1+t)^2 ))dt=∫_0 ^∞ ((8u^6 )/((1+u^2 )^2 ))du. u^6 =(1+u^2 )^3 −3u^2 (1+u^2 )−1=(1+u^2 )^3 −3(1+u^2 )^2 +3(1+u^2 )−1.Hence, A=8∫_0 ^∞ [(1+u^2 )−3+(3/((1+u^2 )))−(1/((1+u^2 )^2 ))]du =8((u^3 /3)−2u+3tan^(−1) (u))∣−∫_0 ^∞ (8/((1+u^2 )^2 ))du Put u=tanθ⇒du=(1+u^2 )dθ B=∫_0 ^(π/2) (8/(1+tan^2 θ))dθ=8∫_0 ^(π/2) cos^2 θdθ =4∫_0 ^(π/2) (1+cos2θ)dθ=4θ+2sin2θ∣_0 ^(π/2) =π The value of this integral isn′t defined](Q108151.png)
$$\left.\mathrm{1}\right)\mathrm{cos}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{4}}>\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{sin}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{4}}\Leftrightarrow\mathrm{cos}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{4}}−\mathrm{sin}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{4}}>\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$ $$\Leftrightarrow\mathrm{cos}\frac{\mathrm{x}}{\mathrm{2}}>\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{cos}\frac{\pi}{\mathrm{4}} \\ $$ $$\Leftrightarrow\mathrm{2}\boldsymbol{\mathrm{k}\pi}−\frac{\boldsymbol{\pi}}{\mathrm{4}}<\boldsymbol{\mathrm{x}}<\frac{\boldsymbol{\pi}}{\mathrm{4}}+\mathrm{2}\boldsymbol{\mathrm{k}\pi} \\ $$ $$\left.\mathrm{2}\right)\mathrm{Put}\:\mathrm{x}=\mathrm{cos}\theta\Rightarrow\mathrm{dx}=−\mathrm{sin}\theta\mathrm{d}\theta \\ $$ $$\mathrm{I}=−\int\sqrt{\frac{\mathrm{1}+\mathrm{cos}\theta}{\mathrm{1}−\mathrm{cos}\theta}}\mathrm{sin}\theta=−\int\sqrt{\frac{\mathrm{2cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}}×\frac{\mathrm{2tan}\frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}\mathrm{d}\theta \\ $$ $$=−\int\mathrm{cot}\frac{\theta}{\mathrm{2}}×\frac{\mathrm{2tan}\frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}\mathrm{d}\theta=−\mathrm{2}\int\frac{\mathrm{d}\theta}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}} \\ $$ $$=−\mathrm{2}\int\mathrm{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\mathrm{d}\theta=−\int\left(\mathrm{1}+\mathrm{cos}\theta\right)\mathrm{d}\theta \\ $$ $$=−\boldsymbol{\theta}−\boldsymbol{\mathrm{sin}\theta}=−\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \boldsymbol{\mathrm{x}}−\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:+\boldsymbol{\mathrm{C}} \\ $$ $$\left.\mathrm{3}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{\mathrm{tanx}}}{\left(\mathrm{cosx}+\mathrm{sinx}\right)^{\mathrm{2}} }\mathrm{dx}=\int\frac{\sqrt{\mathrm{tanx}}}{\mathrm{1}+\mathrm{2sin2x}}\mathrm{dx} \\ $$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{\mathrm{tanx}}}{\mathrm{1}+\frac{\mathrm{2tanx}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx}.\mathrm{Put}\:\mathrm{tanx}=\mathrm{t} \\ $$ $$\Rightarrow\mathrm{dt}=\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dx},\frac{\sqrt{\mathrm{tanx}}}{\mathrm{1}+\frac{\mathrm{2tanx}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}=\frac{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\sqrt{\mathrm{t}}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} } \\ $$ $$\Rightarrow\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} \sqrt{\mathrm{t}}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} }\mathrm{dt}=\int_{\mathrm{0}} ^{\infty} \frac{\left[\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} −\mathrm{2t}\right]^{\mathrm{2}} \sqrt{\mathrm{t}}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} } \\ $$ $$=\int_{\mathrm{0}} ^{\infty} \left[\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} −\mathrm{4t}+\frac{\mathrm{4t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} }\right]\sqrt{\mathrm{t}}\:\mathrm{dt} \\ $$ $$=\int_{\mathrm{0}} ^{\infty} \left[\left(\mathrm{1}−\mathrm{2t}+\mathrm{t}^{\mathrm{2}} \right)\sqrt{\mathrm{t}}\:+\frac{\mathrm{4t}^{\mathrm{2}} \sqrt{\mathrm{t}}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} }\right]\mathrm{dt} \\ $$ $$=\left(\frac{\mathrm{2}}{\mathrm{3}}\mathrm{t}^{\mathrm{3}/\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{5}}\mathrm{t}^{\mathrm{5}/\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{7}}\mathrm{t}^{\mathrm{7}/\mathrm{2}} \right)+\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4t}^{\mathrm{2}} \sqrt{\mathrm{t}}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$ $$\mathrm{Put}\:\sqrt{\mathrm{t}}=\mathrm{u}\Rightarrow\mathrm{du}=\left(\mathrm{1}/\mathrm{2}\sqrt{\mathrm{t}}\right)\mathrm{dt}\Rightarrow\mathrm{dt}=\mathrm{2udu} \\ $$ $$\mathrm{A}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4t}^{\mathrm{2}} \sqrt{\mathrm{t}}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} }\mathrm{dt}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{8u}^{\mathrm{6}} }{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{du}. \\ $$ $$\mathrm{u}^{\mathrm{6}} =\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{3}} −\mathrm{3u}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)−\mathrm{1}=\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$ $$−\mathrm{3}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{3}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)−\mathrm{1}.\mathrm{Hence}, \\ $$ $$\mathrm{A}=\mathrm{8}\int_{\mathrm{0}} ^{\infty} \left[\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)−\mathrm{3}+\frac{\mathrm{3}}{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}−\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\right]\mathrm{du} \\ $$ $$=\mathrm{8}\left(\frac{\mathrm{u}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2u}+\mathrm{3tan}^{−\mathrm{1}} \left(\mathrm{u}\right)\right)\mid−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{8}}{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{du} \\ $$ $$\mathrm{Put}\:\mathrm{u}=\mathrm{tan}\theta\Rightarrow\mathrm{du}=\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\mathrm{d}\theta \\ $$ $$\mathrm{B}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{8}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}\mathrm{d}\theta=\mathrm{8}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$ $$=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\mathrm{cos2}\theta\right)\mathrm{d}\theta=\mathrm{4}\theta+\mathrm{2sin2}\theta\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\pi \\ $$ $$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{integral}}\:\boldsymbol{\mathrm{isn}}'\boldsymbol{\mathrm{t}}\:\boldsymbol{\mathrm{defined}} \\ $$
Commented bymathdave last updated on 15/Aug/20
$${answer}\:{for}\:{question}\:\mathrm{3}\:{is}\:\frac{\pi}{\mathrm{2}} \\ $$
Commented bybemath last updated on 15/Aug/20
$${thank}\:{you}\:{all}\:{sir} \\ $$
Answered by Dwaipayan Shikari last updated on 15/Aug/20
$$\left.\mathrm{2}\right)\int\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx} \\ $$ $$\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$ $${sin}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dt} \\ $$ $${sin}^{−\mathrm{1}} {x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+{C} \\ $$
Answered by john santu last updated on 15/Aug/20
![((⋇JS⋇)/♥) I=∫_0 ^(π/2) ((√(tan x))/(cos^2 x(1+tan x)^2 )) dx I=∫_0 ^(π/2) (((√(tan x)) sec^2 x dx)/((1+tan x)^2 )) [ tan x=b] I=∫_0 ^∞ ((√b)/((1+b)^2 )) db [ set b=(q/(1−q)) ] I=∫_0 ^1 q^(1/2) (1−q)^(−1/2) dq Recall betha function ∫_0 ^1 x^(m−1) (1−x)^(n−1) dx=B(m,n) = ((Γ(m)Γ(n))/(Γ(m+n))) . hence we obtain m=(3/2), n=(1/2). I=(((1/2)Γ((1/2)).Γ((1/2)))/(1!))= (1/2)((√π))^2 =(π/2)](Q108168.png)
$$\:\:\:\:\:\:\:\frac{\divideontimes{JS}\divideontimes}{\heartsuit} \\ $$ $${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\sqrt{\mathrm{tan}\:{x}}}{\mathrm{cos}\:^{\mathrm{2}} {x}\left(\mathrm{1}+\mathrm{tan}\:{x}\right)^{\mathrm{2}} }\:{dx} \\ $$ $${I}=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\sqrt{\mathrm{tan}\:{x}}\:\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}}{\left(\mathrm{1}+\mathrm{tan}\:{x}\right)^{\mathrm{2}} }\:\:\left[\:\mathrm{tan}\:{x}={b}\right] \\ $$ $${I}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\sqrt{{b}}}{\left(\mathrm{1}+{b}\right)^{\mathrm{2}} }\:{db}\:\left[\:{set}\:{b}=\frac{{q}}{\mathrm{1}−{q}}\:\right] \\ $$ $${I}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:{q}^{\mathrm{1}/\mathrm{2}} \left(\mathrm{1}−{q}\right)^{−\mathrm{1}/\mathrm{2}} \:{dq}\: \\ $$ $${Recall}\:{betha}\:{function}\: \\ $$ $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{{m}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}−\mathrm{1}} \:{dx}={B}\left({m},{n}\right) \\ $$ $$=\:\frac{\Gamma\left({m}\right)\Gamma\left({n}\right)}{\Gamma\left({m}+{n}\right)}\:.\:{hence}\:{we}\:{obtain}\: \\ $$ $${m}=\frac{\mathrm{3}}{\mathrm{2}},\:{n}=\frac{\mathrm{1}}{\mathrm{2}}. \\ $$ $${I}=\frac{\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right).\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{1}!}=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\pi}\right)^{\mathrm{2}} =\frac{\pi}{\mathrm{2}} \\ $$
Commented bymathdave last updated on 15/Aug/20
$${brilliant}\:{solution} \\ $$
Commented bymohammad17 last updated on 15/Aug/20
$${sir}\:{can}\:{you}\:{exactly}\:{how}\:\left({m}=\frac{\mathrm{3}}{\mathrm{2}},{n}=\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$