Integration Questions

Question Number 132534 by liberty last updated on 15/Feb/21

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}\:\:\rightarrow\mathrm{diverges}\:\mathrm{or}\:\mathrm{converges}? \\$$

Commented byMJS_new last updated on 15/Feb/21

$$\mathrm{we}\:\mathrm{only}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{if}\:\mathrm{it}\:\mathrm{converges}\:\mathrm{or}\:\mathrm{not} \\$$ $$\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:{x}}\leqslant\mathrm{1}\:\Rightarrow\:\mathrm{converges} \\$$

Answered by Olaf last updated on 15/Feb/21

$$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\mathrm{sin}{x}} \\$$ $$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\mathrm{with}\:{t}\:=\:\mathrm{tan}\frac{{x}}{\mathrm{2}} \\$$ $$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }.\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\$$ $$\Omega\:=\:−\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{1}+{t}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\:−\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}\frac{{x}}{\mathrm{2}}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$ $$\Omega\:=\:−\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right]\:=\:\mathrm{1} \\$$

Answered by EDWIN88 last updated on 15/Feb/21

$$\:\mathrm{We}\:\mathrm{want}\:\mathrm{to}\:\mathrm{evaluate}\:\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+\mathrm{sin}\:{x}}\:. \\$$ $$\:\mathrm{I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{2sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \\$$ $$\mathrm{Wierrstrass}\:\mathrm{Substitution} \\$$ $$\mathrm{let}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\varnothing\:\Rightarrow\begin{cases}{\mathrm{x}=\frac{\pi}{\mathrm{2}}\rightarrow\varnothing=\mathrm{1}}\\{\mathrm{x}=\mathrm{0}\rightarrow\varnothing=\mathrm{0}}\end{cases} \\$$ $$\:\:\mathrm{dx}\:=\:\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{d}\varnothing \\$$ $$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\left(\frac{\varnothing}{\mathrm{1}+\varnothing^{\mathrm{2}} }\right)}.\left(\frac{\mathrm{2}}{\mathrm{1}+\varnothing^{\mathrm{2}} }\right)\mathrm{d}\varnothing \\$$ $$\mathrm{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}}{\left(\mathrm{1}+\varnothing\right)^{\mathrm{2}} }\mathrm{d}\varnothing\:=\:−\mathrm{2}\:\left[\frac{\mathrm{1}}{\mathrm{1}+\varnothing}\:\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$ $$\mathrm{I}=−\mathrm{2}\:\left[\:\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\:\right]=\:\mathrm{1}\:\left(\mathrm{converges}\:\right)\: \\$$

Commented byliberty last updated on 15/Feb/21

Answered by Ar Brandon last updated on 15/Feb/21

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−\mathrm{sinx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}=\left[\mathrm{tanx}−\frac{\mathrm{1}}{\mathrm{cosx}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{1} \\$$

Commented byEDWIN88 last updated on 15/Feb/21

$$\:\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\pi}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{cos}\:\frac{\pi}{\mathrm{2}}}\:=?\:\Rightarrow\frac{\mathrm{1}}{\infty}−\frac{\mathrm{1}}{\mathrm{0}}\:?\: \\$$

Commented byAr Brandon last updated on 15/Feb/21

$$\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left\{\mathrm{tanx}−\frac{\mathrm{1}}{\mathrm{cosx}}\right\}=\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left\{\frac{\mathrm{sinx}−\mathrm{1}}{\mathrm{cosx}}\right\}=\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left\{\frac{\mathrm{cosx}}{−\mathrm{sinx}}\right\}=\mathrm{0} \\$$

Commented byAr Brandon last updated on 15/Feb/21

$$\mathrm{tan}\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{cos}\frac{\pi}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{cos}\frac{\pi}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{cos}\frac{\pi}{\mathrm{2}}}=\mathrm{0} \\$$