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Question Number 43625 by peter frank last updated on 12/Sep/18

∫_( 0) ^(π/4)   ((√(tan x))/(sin x cos x)) dx = 2

$$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\:\:\frac{\sqrt{\mathrm{tan}\:{x}}}{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}\:{dx}\:=\:\mathrm{2} \\ $$

Answered by last updated on 13/Sep/18

∫_0 ^(Π/4) (((√(tanx)) )/(tanx.cos^2 x))dx  ∫_0 ^(Π/4) ((sec^2 x)/(√(tanx))) dx  t^2 =tanx  2tdt=sec^2 xdx  ∫_0 ^1 ((2tdt)/t)  2×∣t∣_0 ^1 =2

$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} \frac{\sqrt{{tanx}}\:}{{tanx}.{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} \frac{{sec}^{\mathrm{2}} {x}}{\sqrt{{tanx}}}\:{dx} \\ $$$${t}^{\mathrm{2}} ={tanx} \\ $$$$\mathrm{2}{tdt}={sec}^{\mathrm{2}} {xdx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{tdt}}{{t}} \\ $$$$\mathrm{2}×\mid{t}\mid_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{2} \\ $$

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