Integration Questions

Question Number 106964 by Ar Brandon last updated on 08/Aug/20

$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx} \\$$

Commented by Her_Majesty last updated on 08/Aug/20

$${I}\:{believe}\:{this}\:{integral}\:{is}\:{divergent} \\$$$${must}\:{sleep}\:{now}\:{to}\:{keep}\:{my}\:{majestic}\:{beauty} \\$$$${will}\:{try}\:{to}\:{show}\:{later} \\$$

Commented by Dwaipayan Shikari last updated on 08/Aug/20

$$\mathrm{probably} \\$$

Commented by Dwaipayan Shikari last updated on 08/Aug/20

Commented by Ar Brandon last updated on 08/Aug/20

��Alright her majesty ��

Answered by bemath last updated on 08/Aug/20

$$\:\:\:\:^{@{bemath}@} \\$$$${let}\:{u}\:=\:\mathrm{tan}\:{x}\:\rightarrow\begin{cases}{{du}=\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}}\\{\rightarrow\begin{cases}{{x}=\mathrm{0}\rightarrow{u}=\mathrm{0}}\\{{x}=\pi\rightarrow{u}=\mathrm{0}}\end{cases}}\end{cases} \\$$$${I}=\underset{\mathrm{0}} {\overset{\mathrm{0}} {\int}}\:\frac{{du}}{\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}\:=\:\mathrm{0} \\$$

Answered by 1549442205PVT last updated on 08/Aug/20

$$\mathrm{I}=\int_{\mathrm{0}} ^{\:\pi} \frac{\mathrm{d}\left(\mathrm{tanx}\right)}{\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}\underset{\mathrm{tanx}=\mathrm{u}} {=\:}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{du}}{\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }}=\mathrm{sin}^{−\mathrm{1}} \mathrm{u}\mid_{\mathrm{0}} ^{\mathrm{0}} \\$$$$=\mathrm{0}−\mathrm{0}=\mathrm{0} \\$$

Answered by Dwaipayan Shikari last updated on 08/Aug/20

$$\int_{\mathrm{0}} ^{\pi} \frac{{dt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\:\:\:\:\:\:\:\:\:{tanx}={t}\:\:\:\:\:,{sec}^{\mathrm{2}} {x}=\frac{{dt}}{{dx}} \\$$$$\left[{sin}^{−\mathrm{1}} \left({tanx}\right)\right]_{\mathrm{0}} ^{\pi} =\mathrm{0} \\$$

Commented by Ar Brandon last updated on 08/Aug/20

$$\mathrm{Thank}\:\mathrm{y}'\mathrm{all}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{needed}\:\mathrm{to}\:\mathrm{verify}\:\mathrm{something}. \\$$$$\mathrm{And}\:\mathrm{you}'\mathrm{ve}\:\mathrm{been}\:\mathrm{of}\:\mathrm{great}\:\mathrm{help}. \\$$

Answered by Ar Brandon last updated on 08/Aug/20

$$\mathcal{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx}+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx} \\$$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx}=\mathrm{0} \\$$

Answered by Her_Majesty last updated on 08/Aug/20

$${y}=\frac{{sec}^{\mathrm{2}} {x}}{\sqrt{\mathrm{1}−{tan}^{\mathrm{2}} {x}}}=\frac{\mathrm{1}}{\mid{cosx}\mid\sqrt{{cos}\mathrm{2}{x}}} \\$$$${for}\:\mathrm{0}\leqslant{x}\leqslant\pi\:\:{y}\:{is}\:{defined}\:{for}\:\mathrm{0}\leqslant{x}<\frac{\pi}{\mathrm{4}}\:{and} \\$$$$\frac{\mathrm{3}\pi}{\mathrm{4}}<{y}\leqslant\pi\:\:\Rightarrow\:{the}\:{integral}\:{only}\:{exists}\:{if}\:{the} \\$$$${imaginary}/{complex}\:{parts}\:{cancel}\:{out} \\$$$${but}\:{let}'{s}\:{try} \\$$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dx}}{\mid{cosx}\mid\sqrt{{cos}\mathrm{2}{x}}}=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{dx}}{{cosx}\sqrt{{cos}\mathrm{2}{x}}} \\$$$${with}\:{t}={tanx}\:{we}\:{get} \\$$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{dx}}{{cosx}\sqrt{{cos}\mathrm{2}{x}}}=\underset{\mathrm{0}} {\overset{+\infty} {\int}}\frac{{dt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\left[{arcsin}\left({t}\right)\right]_{\mathrm{0}} ^{+\infty} = \\$$$$=\underset{{t}\rightarrow+\infty} {{lim}arcsin}\left({t}\right)\:{which}\:{is}\:\frac{\pi}{\mathrm{2}}−{i}\infty\:\left({try}\:{some}\right. \\$$$${values}\:{like}\:{t}=\mathrm{10}^{\mathrm{10}} ,\:\mathrm{10}^{\mathrm{20}} ... \\$$$$\Rightarrow\:{the}\:{integral}\:{is}\:{divergent} \\$$

Commented by Ar Brandon last updated on 08/Aug/20

OK Sir, thanks