Question Number 223204 by Nadirhashim last updated on 17/Jul/25
$$\:\:\:\underset{\frac{\pi}{\mathrm{6}}} {\overset{\frac{\pi}{\mathrm{3}}} {\int}}\frac{\boldsymbol{{dx}}}{\mathrm{1}+\sqrt{\boldsymbol{{tanx}}}}\:=…? \\ $$
Answered by som(math1967) last updated on 17/Jul/25
![I=∫_((π/6) ) ^(π/3) (dx/(1+(√(tan((π/3)+(π/6)−x))))) I=∫_((π/6) ) ^(π/3) (dx/(1+(√(cotx))))=∫_((π/6) ) ^(π/3) (((√(tanx))dx)/(1+(√(tanx)))) 2I=∫_((π/6) ) ^(π/3) (dx/(1+(√(tanx)))) +∫_((π/6) ) ^(π/3) (((√(tanx))dx)/(1+(√(tanx)))) ⇒2I=∫_((π/6) ) ^(π/3) ((1+(√(tanx))dx)/(1+(√(tanx))))=∫_((π/6) ) ^(π/3) dx ⇒2I=[x]_(π/6) ^(π/3) =(π/3)−(π/6) ∴ I=(π/(12))](https://www.tinkutara.com/question/Q223206.png)
$$\:\:\:{I}=\int_{\frac{\pi}{\mathrm{6}}\:} ^{\frac{\pi}{\mathrm{3}}} \frac{{dx}}{\mathrm{1}+\sqrt{{tan}\left(\frac{\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{6}}−{x}\right)}} \\ $$$${I}=\int_{\frac{\pi}{\mathrm{6}}\:} ^{\frac{\pi}{\mathrm{3}}} \frac{{dx}}{\mathrm{1}+\sqrt{{cotx}}}=\int_{\frac{\pi}{\mathrm{6}}\:} ^{\frac{\pi}{\mathrm{3}}} \frac{\sqrt{{tanx}}{dx}}{\mathrm{1}+\sqrt{{tanx}}} \\ $$$$\mathrm{2}{I}=\int_{\frac{\pi}{\mathrm{6}}\:} ^{\frac{\pi}{\mathrm{3}}} \frac{{dx}}{\mathrm{1}+\sqrt{{tanx}}}\:+\int_{\frac{\pi}{\mathrm{6}}\:} ^{\frac{\pi}{\mathrm{3}}} \frac{\sqrt{{tanx}}{dx}}{\mathrm{1}+\sqrt{{tanx}}} \\ $$$$\Rightarrow\mathrm{2}{I}=\int_{\frac{\pi}{\mathrm{6}}\:} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{1}+\sqrt{{tanx}}{dx}}{\mathrm{1}+\sqrt{{tanx}}}=\int_{\frac{\pi}{\mathrm{6}}\:} ^{\frac{\pi}{\mathrm{3}}} {dx} \\ $$$$\Rightarrow\mathrm{2}{I}=\left[{x}\right]_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} =\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{6}} \\ $$$$\therefore\:{I}=\frac{\pi}{\mathrm{12}} \\ $$$$ \\ $$