Question Number 227128 by Spillover last updated on 01/Jan/26
Answered by som(math1967) last updated on 01/Jan/26
![∫_(0 ) ^(π/2) ((√(tanx))/(1+(√(tanx))))dx I=∫_0 ^(π/2) (((√(tan((π/2)−x)))dx)/(1+(√(tan((π/2)−x))))) I=∫_0 ^(π/2) (((√(cotx))dx)/(1+(√(cotx))))=∫_0 ^(π/2) (((√(cotx))dx)/(1+(1/( (√(tanx)))))) I=∫_0 ^(π/2) (dx/(1+(√(tanx)))) 2I=∫_0 ^(π/2) ((1+(√(tanx))dx)/(1+(√(tanx))))=∫_0 ^(π/2) dx 2I=[x]_0 ^(π/2) ⇒I=(π/4)](https://www.tinkutara.com/question/Q227129.png)
$$\int_{\mathrm{0}\:\:} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{{tanx}}}{\mathrm{1}+\sqrt{{tanx}}}{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{{tan}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx}}{\mathrm{1}+\sqrt{{tan}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{{cotx}}{dx}}{\mathrm{1}+\sqrt{{cotx}}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{{cotx}}{dx}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{tanx}}}} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\sqrt{{tanx}}} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+\sqrt{{tanx}}{dx}}{\mathrm{1}+\sqrt{{tanx}}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\ $$$$\mathrm{2}{I}=\left[{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \Rightarrow{I}=\frac{\pi}{\mathrm{4}} \\ $$
Commented by Spillover last updated on 02/Jan/26
$${tbank}\:{you} \\ $$