Question Number 122587 by bramlexs22 last updated on 18/Nov/20
$$\:\:\underset{\mathrm{1}} {\overset{\mathrm{16}} {\int}}\:\mathrm{arctan}\:\sqrt{\sqrt{{x}}\:−\mathrm{1}}\:{dx}\: \\ $$$$\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{sin}\:\left(\mathrm{2}{x}\right)\:\mathrm{arctan}\:\left(\mathrm{sin}\:{x}\right)\:{dx}\: \\ $$
Answered by MJS_new last updated on 18/Nov/20
![∫sin 2x arctan sin x dx= [by parts] =−(1/2)cos 2x arctan sin x +(1/2)∫((cos x cos 2x)/(1+sin^2 x))dx= (1/2)∫((cos x cos 2x)/(1+sin^2 x))dx= [t=sin x → dx=(dt/(cos x))] =−(1/2)∫((2t^2 −1)/(t^2 +1))dt=−∫dt+(3/2)∫(dt/(t^2 +1))= =−t+(3/2)arctan t =−sin x +(3/2)arctan sin x =−sin x +((3−cos 2x)/2)arctan sin x +C ⇒ answer is (π/2)−1](Q122594.png)
$$\int\mathrm{sin}\:\mathrm{2}{x}\:\mathrm{arctan}\:\mathrm{sin}\:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\:\mathrm{arctan}\:\mathrm{sin}\:{x}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{t}=\mathrm{sin}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{cos}\:{x}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=−\int{dt}+\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=−{t}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{arctan}\:{t}\:=−\mathrm{sin}\:{x}\:+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{arctan}\:\mathrm{sin}\:{x} \\ $$$$ \\ $$$$=−\mathrm{sin}\:{x}\:+\frac{\mathrm{3}−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}\mathrm{arctan}\:\mathrm{sin}\:{x}\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\pi}{\mathrm{2}}−\mathrm{1} \\ $$
Answered by liberty last updated on 18/Nov/20
![(1) ∫_1 ^(16) arctan (√((√x)−1)) dx letting (√((√x) −1)) = ♭ → { ((♭=0)),((♭=(√3))) :} ⇒(√x) = ♭^2 +1 ⇒x=(♭^2 +1)^2 ⇒ dx = 4♭(♭^2 +1) d♭ R=∫_0 ^(√3) (4♭^3 +4♭) arctan ♭ d♭ [ D.I method ] R = [ (♭^4 +2♭^2 ) arctan ♭ ]_0 ^(√3) −∫_0 ^(√3) [((♭^4 +2♭^2 )/(1+♭^2 ))] d♭ R= 5π−∫_0 ^(√3) (((1+♭^2 )^2 −1)/(1+♭^2 )) d♭ R=5π−∫_0 ^(√3) (1+♭^2 )d♭+∫_0 ^(√3) (d♭/(1+♭^2 )) R=5π−(♭+(1/3)♭^3 )_0 ^(√3) +arctan (♭)]_0 ^(√3) R=5π−2(√3)+(π/3) = ((16π)/3)−2(√3). ▲](Q122595.png)
$$\left(\mathrm{1}\right)\:\underset{\mathrm{1}} {\overset{\mathrm{16}} {\int}}\:\mathrm{arctan}\:\sqrt{\sqrt{\mathrm{x}}−\mathrm{1}}\:\mathrm{dx}\: \\ $$$$\mathrm{letting}\:\sqrt{\sqrt{\mathrm{x}}\:−\mathrm{1}}\:=\:\flat\:\rightarrow\begin{cases}{\flat=\mathrm{0}}\\{\flat=\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\:\Rightarrow\sqrt{\mathrm{x}}\:=\:\flat^{\mathrm{2}} +\mathrm{1}\:\Rightarrow\mathrm{x}=\left(\flat^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{dx}\:=\:\mathrm{4}\flat\left(\flat^{\mathrm{2}} +\mathrm{1}\right)\:\mathrm{d}\flat\: \\ $$$$\mathrm{R}=\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{3}}} {\int}}\:\left(\mathrm{4}\flat^{\mathrm{3}} +\mathrm{4}\flat\right)\:\mathrm{arctan}\:\flat\:\mathrm{d}\flat\: \\ $$$$\left[\:\mathrm{D}.\mathrm{I}\:\mathrm{method}\:\right]\: \\ $$$$\mathrm{R}\:=\:\left[\:\left(\flat^{\mathrm{4}} +\mathrm{2}\flat^{\mathrm{2}} \right)\:\mathrm{arctan}\:\flat\:\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} −\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{3}}} {\int}}\:\left[\frac{\flat^{\mathrm{4}} +\mathrm{2}\flat^{\mathrm{2}} }{\mathrm{1}+\flat^{\mathrm{2}} }\right]\:\mathrm{d}\flat \\ $$$$\mathrm{R}=\:\mathrm{5}\pi−\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{3}}} {\int}}\:\frac{\left(\mathrm{1}+\flat^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+\flat^{\mathrm{2}} }\:\mathrm{d}\flat \\ $$$$\mathrm{R}=\mathrm{5}\pi−\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{3}}} {\int}}\left(\mathrm{1}+\flat^{\mathrm{2}} \right)\mathrm{d}\flat+\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{3}}} {\int}}\:\frac{\mathrm{d}\flat}{\mathrm{1}+\flat^{\mathrm{2}} } \\ $$$$\left.\mathrm{R}=\mathrm{5}\pi−\left(\flat+\frac{\mathrm{1}}{\mathrm{3}}\flat^{\mathrm{3}} \right)_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} +\mathrm{arctan}\:\left(\flat\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \\ $$$$\mathrm{R}=\mathrm{5}\pi−\mathrm{2}\sqrt{\mathrm{3}}+\frac{\pi}{\mathrm{3}}\:=\:\frac{\mathrm{16}\pi}{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{3}}.\:\blacktriangle \\ $$
Answered by MJS_new last updated on 18/Nov/20
![∫arctan (√((√x)−1)) dx= [by parts] =xarctan (√((√x)−1)) −(1/4)∫(dx/( (√((√x)−1))))= −(1/4)∫(dx/( (√((√x)−1))))= [t=(√((√x)−1)) → dx=4(√x)(√((√x)−1))dt] =−∫t^2 +1dt=−(t^3 /3)−t= =−(((2+(√x))(√((√x)−1)))/3) =xarctan (√((√x)−1)) −(((2+(√x))(√((√x)−1)))/3)+C ⇒ answer is ((16π)/3)−2(√3)](Q122598.png)
$$\int\mathrm{arctan}\:\sqrt{\sqrt{{x}}−\mathrm{1}}\:{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$={x}\mathrm{arctan}\:\sqrt{\sqrt{{x}}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{\:\sqrt{\sqrt{{x}}−\mathrm{1}}}= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{\:\sqrt{\sqrt{{x}}−\mathrm{1}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{t}=\sqrt{\sqrt{{x}}−\mathrm{1}}\:\rightarrow\:{dx}=\mathrm{4}\sqrt{{x}}\sqrt{\sqrt{{x}}−\mathrm{1}}{dt}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\int{t}^{\mathrm{2}} +\mathrm{1}{dt}=−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−{t}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\left(\mathrm{2}+\sqrt{{x}}\right)\sqrt{\sqrt{{x}}−\mathrm{1}}}{\mathrm{3}} \\ $$$$ \\ $$$$={x}\mathrm{arctan}\:\sqrt{\sqrt{{x}}−\mathrm{1}}\:−\frac{\left(\mathrm{2}+\sqrt{{x}}\right)\sqrt{\sqrt{{x}}−\mathrm{1}}}{\mathrm{3}}+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{16}\pi}{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{3}} \\ $$
Commented by liberty last updated on 18/Nov/20
$$\mathrm{why}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{not}\:\mathrm{same}\:\mathrm{with}\:\mathrm{your} \\ $$$$\mathrm{answer}\:\mathrm{sir}?\:\mathrm{what}\:\mathrm{wrong}? \\ $$
Commented by MJS_new last updated on 18/Nov/20
$$\mathrm{I}\:\mathrm{made}\:\mathrm{a}\:\mathrm{mistake},\:\mathrm{I}'\mathrm{ll}\:\mathrm{correct}\:\mathrm{it} \\ $$
Answered by Dwaipayan Shikari last updated on 18/Nov/20
![∫_1 ^(16) tan^(−1) ((√((√x)−1)))dx x=t^2 ⇒1=2t(dt/dx) ∫_1 ^4 2t tan^(−1) ((√(t−1)))dt t−1=u^2 ⇒1=2u(du/dt) 4∫_0 ^(√3) u(u^2 +1)tan^(−1) udu ∫u^3 tan^(−1) udu=(u^4 /4)tan^(−1) u−(u^3 /(12))+(u/4)−(1/4)tan^(−1) u ∫utan^(−1) udu=(u^2 /2)tan^(−1) u−(u/2)+(1/2)tan^(−1) u So it becomes 4[(9/4).(π/3)−((3(√3))/(12))+((√3)/4)−(π/(12))]+4[(3/2).(π/3)−((√3)/2)+(π/6)] 5π+(π/3)−2(√3)=((16π)/3)−2(√3)](Q122613.png)
$$\int_{\mathrm{1}} ^{\mathrm{16}} {tan}^{−\mathrm{1}} \left(\sqrt{\sqrt{{x}}−\mathrm{1}}\right){dx}\:\:\:\:\:\:\:\:\:\:{x}={t}^{\mathrm{2}} \Rightarrow\mathrm{1}=\mathrm{2}{t}\frac{{dt}}{{dx}} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{4}} \mathrm{2}{t}\:\:{tan}^{−\mathrm{1}} \left(\sqrt{{t}−\mathrm{1}}\right){dt}\:\:\:\:\:\:\:{t}−\mathrm{1}={u}^{\mathrm{2}} \Rightarrow\mathrm{1}=\mathrm{2}{u}\frac{{du}}{{dt}} \\ $$$$\mathrm{4}\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} {u}\left({u}^{\mathrm{2}} +\mathrm{1}\right){tan}^{−\mathrm{1}} {udu} \\ $$$$\int{u}^{\mathrm{3}} {tan}^{−\mathrm{1}} {udu}=\frac{{u}^{\mathrm{4}} }{\mathrm{4}}{tan}^{−\mathrm{1}} {u}−\frac{{u}^{\mathrm{3}} }{\mathrm{12}}+\frac{{u}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}{tan}^{−\mathrm{1}} {u} \\ $$$$\int{utan}^{−\mathrm{1}} {udu}=\frac{{u}^{\mathrm{2}} }{\mathrm{2}}{tan}^{−\mathrm{1}} {u}−\frac{{u}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} {u} \\ $$$${So}\:{it}\:{becomes} \\ $$$$\mathrm{4}\left[\frac{\mathrm{9}}{\mathrm{4}}.\frac{\pi}{\mathrm{3}}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{12}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}−\frac{\pi}{\mathrm{12}}\right]+\mathrm{4}\left[\frac{\mathrm{3}}{\mathrm{2}}.\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}\right] \\ $$$$\mathrm{5}\pi+\frac{\pi}{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{3}}=\frac{\mathrm{16}\pi}{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{3}} \\ $$
Answered by Bird last updated on 18/Nov/20
![I =∫_1 ^(16) srctan((√((√x)−1)))dx changrment (√((√x)−1))=t give (√x)−1=t^(2 ) ⇒(√x)=t^2 +1 ⇒ x=(t^2 +1)^2 ⇒ I =∫_0 ^(√3) arctan(t)4t(t^2 +1)dt =4 ∫_0 ^(√3) (t^3 +t)arcctant dt =4( ((t^4 /4)+(t^2 /2))arctsnt]_0 ^(√3) −∫_0 ^(√3) ((t^4 /4)+(t^2 /2))(dt/(1+t^2 ))} =4{( (9/4)+(3/2))(π/3)}−∫_0 ^(√3) ((t^4 +2t^2 )/(t^2 +1))dt =(9+6)(π/3)−∫_0 ^(√3) ((t^4 +2t^2 )/(t^2 +1))dt =5π−H H =∫_0 ^(√3) ((t^2 (t^2 +1)+t^2 )/(t^2 +1))dt =∫_0 ^(√3) t^2 dt +∫_0 ^(√3) ((t^2 +1−1)/(t^2 +1))dt =[(t^3 /3)]_0 ^(√3) +(√3)−[arctan(t)]_0 ^(√3) =(√3)+(√3)−(π/3)=2(√3)−(π/3) ⇒ I =5π−2(√3)+(π/3) =((16π)/3)−2(√3)](Q122687.png)
$${I}\:=\int_{\mathrm{1}} ^{\mathrm{16}} \:{srctan}\left(\sqrt{\sqrt{{x}}−\mathrm{1}}\right){dx}\: \\ $$$${changrment}\:\sqrt{\sqrt{{x}}−\mathrm{1}}={t}\:{give} \\ $$$$\sqrt{{x}}−\mathrm{1}={t}^{\mathrm{2}\:} \:\Rightarrow\sqrt{{x}}={t}^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow \\ $$$${x}=\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} {arctan}\left({t}\right)\mathrm{4}{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right){dt} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \left({t}^{\mathrm{3}} \:+{t}\right){arcctant}\:{dt} \\ $$$$=\mathrm{4}\left(\:\:\left(\frac{{t}^{\mathrm{4}} }{\mathrm{4}}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right){arctsnt}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \\ $$$$\left.−\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \left(\frac{{t}^{\mathrm{4}} }{\mathrm{4}}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right)\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\right\} \\ $$$$=\mathrm{4}\left\{\left(\:\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}\right)\frac{\pi}{\mathrm{3}}\right\}−\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \frac{{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$$=\left(\mathrm{9}+\mathrm{6}\right)\frac{\pi}{\mathrm{3}}−\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\frac{{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$$=\mathrm{5}\pi−{H} \\ $$$${H}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \frac{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)+{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} {t}^{\mathrm{2}} \:{dt}\:+\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\frac{{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$=\left[\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:+\sqrt{\mathrm{3}}−\left[{arctan}\left({t}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \\ $$$$=\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}−\frac{\pi}{\mathrm{3}}=\mathrm{2}\sqrt{\mathrm{3}}−\frac{\pi}{\mathrm{3}}\:\Rightarrow \\ $$$${I}\:=\mathrm{5}\pi−\mathrm{2}\sqrt{\mathrm{3}}+\frac{\pi}{\mathrm{3}}\:=\frac{\mathrm{16}\pi}{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$ \\ $$