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Question Number 83146 by 09658867628 last updated on 28/Feb/20

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

Answered by Kunal12588 last updated on 28/Feb/20

$${I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \frac{\sqrt{{cot}\:{x}}}{\sqrt{{cot}\:{x}}+\sqrt{{tan}\:{x}}}{dx} \\$$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \frac{\sqrt{{tan}\:{x}}}{\sqrt{{tan}\:{x}}+\sqrt{{cot}\:{x}}}{dx} \\$$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {dx} \\$$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{4}} \\$$$$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\:\frac{\sqrt{\mathrm{cot}\:{x}}}{\sqrt{\mathrm{cot}\:{x}}\:+\:\sqrt{\mathrm{tan}\:{x}}}\:{dx}\:=\frac{\pi}{\mathrm{4}} \\$$$$\\$$

Answered by niroj last updated on 28/Feb/20

$$\:\mathrm{let},\:\mathrm{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\:\sqrt{\mathrm{cot}\:\mathrm{x}}}{\:\sqrt{\mathrm{cot}\:\mathrm{x}}\:+\sqrt{\mathrm{tan}\:\mathrm{x}}}\mathrm{dx}.....\left(\mathrm{i}\right) \\$$$$\:\:\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\sqrt{\:\mathrm{cot}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}}{\left.\:\sqrt{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{2}}\right.}\:−\mathrm{x}\right)+\:\sqrt{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}\:−\mathrm{x}\right)}}\mathrm{dx}\:\:\:\:\:\because\:\int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{xdx}=\int_{\mathrm{0}} ^{\mathrm{a}} \left(\mathrm{a}−\mathrm{x}\right)\mathrm{dx} \\$$$$\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}}{\:\sqrt{\mathrm{tan}\:\mathrm{x}}\:+\sqrt{\mathrm{cot}\:\mathrm{x}}}\mathrm{dx}......\left(\mathrm{ii}\right) \\$$$$\:\:\mathrm{added}\:\left(\mathrm{i}\right)\&\left(\mathrm{ii}\right) \\$$$$\:\:\mathrm{2I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\:\:\frac{\:\:\:\sqrt{\mathrm{cot}\:\mathrm{x}}}{\:\sqrt{\mathrm{cot}\:\mathrm{x}}\:\:+\sqrt{\mathrm{tan}\:\mathrm{x}}}\:+\:\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}}{\:\:\sqrt{\mathrm{tan}\:\mathrm{x}}\:+\sqrt{\mathrm{cot}\:\mathrm{x}}}\right)\mathrm{dx} \\$$$$\:\:\mathrm{2I}=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\:\left(\frac{\:\:\sqrt{\mathrm{cot}\:\mathrm{x}}\:\:+\sqrt{\mathrm{tan}\:\mathrm{x}}}{\:\:\sqrt{\mathrm{cot}\:\mathrm{x}}+\sqrt{\mathrm{tan}\:\mathrm{x}}}\right)\mathrm{dx} \\$$$$\:\:\mathrm{2I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{dx} \\$$$$\:\:\mathrm{2I}=\:\left[\:\mathrm{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$$$\:\:\mathrm{2I}=\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{0}\right) \\$$$$\:\:\:\:\mathrm{2I}=\:\frac{\pi}{\mathrm{2}}\:\Rightarrow\:\mathrm{I}=\:\frac{\pi}{\mathrm{4}}\://. \\$$$$\\$$$$\\$$

Commented by peter frank last updated on 28/Feb/20

$${thank}\:{you}\:{both} \\$$