Integration Questions

Question Number 86313 by john santu last updated on 28/Mar/20

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dx}}{\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}} \\$$

Commented by john santu last updated on 28/Mar/20

$$\frac{\sqrt{\mathrm{sin}\:{x}}}{\sqrt{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}}\:+\:\frac{\sqrt{\mathrm{cos}\:{x}}}{\sqrt{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}}\:=\:\mathrm{1} \\$$$${how}\:? \\$$

Commented by john santu last updated on 28/Mar/20

$$\sqrt{\mathrm{cos}\:{x}}\:+\:\sqrt{\mathrm{sin}\:{x}}\:=\:\sqrt{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:? \\$$

Answered by TANMAY PANACEA. last updated on 28/Mar/20

$${i}\:{think}... \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\frac{\sqrt{{sinx}}}{\sqrt{{cosx}}}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{{cosx}}}{\sqrt{{sinx}}\:+\sqrt{{cosx}}}{dx} \\$$$${using}\:\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{a}} {f}\left({a}−{x}\right){dx} \\$$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}}{\sqrt{{sin}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}\:+\sqrt{{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}}{dx} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{{sinx}}}{\sqrt{{cosx}}\:+\sqrt{{sinx}}}{dx} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{{cosx}}}{\sqrt{{sinx}}\:+\sqrt{{cosx}}}+\frac{\sqrt{{sinx}}}{\sqrt{{cosx}}\:+\sqrt{{sinx}}}{dx} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\$$$${I}=\frac{\pi}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\pi}{\mathrm{4}} \\$$

Commented by jagoll last updated on 28/Mar/20

$$\mathrm{king}\:\mathrm{formula} \\$$

Commented by TANMAY PANACEA. last updated on 28/Mar/20

$${thank}\:{you}\:{sir} \\$$

Commented by john santu last updated on 28/Mar/20

$$\mathrm{it}\:\mathrm{does}\:\mathrm{mean}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{wrong}\:\mathrm{sir}? \\$$

Commented by TawaTawa1 last updated on 28/Mar/20

$$\mathrm{Sir}\:\mathrm{tanmay},\:\mathrm{please}\:\mathrm{help}\:\mathrm{with}\:\:\mathrm{Q86324}. \\$$

Answered by MJS last updated on 28/Mar/20

$$\int\frac{{dx}}{\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}}= \\$$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}{dt}\right] \\$$$$=\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}} \\$$$$\mathrm{and}\:\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy} \\$$

Commented by jagoll last updated on 28/Mar/20

$$=\:\mathrm{2}\:\int\:\frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\: \\$$$$\mathrm{let}\:\mathrm{t}^{\mathrm{2}} −\mathrm{1}\:=\:\mathrm{tan}\:\mathrm{w}\Rightarrow\:\mathrm{2t}\:\mathrm{dt}\:=\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{w}\:\mathrm{dw} \\$$$$=\:\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{w}\:\mathrm{dw}}{\mathrm{sec}\:^{\mathrm{2}} \mathrm{w}}\:=\:\mathrm{w}\:+\:\mathrm{c} \\$$$$=\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)\:+\:\mathrm{c} \\$$$$=\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}\:\left(\mathrm{x}\right)\right)\:+\:\mathrm{c} \\$$$$\mathrm{it}\:\mathrm{correct}\:\mathrm{sir}? \\$$

Commented by MJS last updated on 28/Mar/20

$$\mathrm{no} \\$$$${w}=\mathrm{arctan}\:\left({t}^{\mathrm{2}} −\mathrm{1}\right)\:\rightarrow\:{dt}=\frac{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}}{\mathrm{2}{t}}{dw} \\$$$$\Rightarrow\:\int\frac{{dw}}{\sqrt{\mathrm{1}+\mathrm{tan}\:{w}}}\:\mathrm{so}\:\mathrm{you}'\mathrm{re}\:\mathrm{back}\:\mathrm{at}\:\mathrm{the}\:\mathrm{start} \\$$

Commented by MJS last updated on 28/Mar/20

$$\mathrm{we}\:\mathrm{must}\:\mathrm{decompose} \\$$$$\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}= \\$$$$=\int\frac{{at}+{b}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{t}+\sqrt{\mathrm{2}}}{dt}+\int\frac{{ct}+{d}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}{t}+\sqrt{\mathrm{2}}}}{dt}= \\$$$$=−\frac{\sqrt{−\mathrm{1}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\int\frac{{t}−\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{t}+\sqrt{\mathrm{2}}}{dt}+\frac{\sqrt{−\mathrm{1}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\int\frac{{t}+\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}{t}+\sqrt{\mathrm{2}}}{dt} \\$$$$\mathrm{now}\:\mathrm{use}\:\mathrm{formula} \\$$

Commented by jagoll last updated on 28/Mar/20

$$\mathrm{waw}...\mathrm{yes}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{try}\:\mathrm{solve}\:\mathrm{it} \\$$