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Question Number 58168 by maxmathsup by imad last updated on 19/Apr/19

find ∫    ((√(tanx))/(sin(2x)))dx

$${find}\:\int\:\:\:\:\frac{\sqrt{{tanx}}}{{sin}\left(\mathrm{2}{x}\right)}{dx} \\ $$

Answered by tanmay last updated on 19/Apr/19

∫(((√(tanx)) )/(2tanx))×(1+tan^2 x)dx  (1/2)∫((sec^2 x)/(√(tanx)))dx  (1/2)∫(tanx)^((−1)/2) d(tanx)  (1/2)×(((tanx)^(((−1)/2)+1) )/(((−1)/2)+1))+c  (√(tanx)) +c

$$\int\frac{\sqrt{{tanx}}\:}{\mathrm{2}{tanx}}×\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{sec}^{\mathrm{2}} {x}}{\sqrt{{tanx}}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\left({tanx}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} {d}\left({tanx}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({tanx}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}+\mathrm{1}} }{\frac{−\mathrm{1}}{\mathrm{2}}+\mathrm{1}}+{c} \\ $$$$\sqrt{{tanx}}\:+{c} \\ $$

Commented by maxmathsup by imad last updated on 19/Apr/19

thanks sir

$${thanks}\:{sir} \\ $$

Answered by malwaan last updated on 19/Apr/19

∫((√(tanx))/(2sinxcosx))dx  =∫((√(tanx))/(2((sinx)/(cosx))×cos^2 x)) dx  =(1/2)∫(( sec^2 x(√(tanx)))/(tanx)) dx  =(1/2)∫((sec^2 x)/(√(tanx))) dx=(1/2)∫(tanx)^(−(1/2)) sec^2 xdx  =(1/2)(((tanx)^(1/2) )/(1/2)) +c=(√(tanx)) +c

$$\int\frac{\sqrt{\mathrm{tan}{x}}}{\mathrm{2}{sinxcosx}}{dx} \\ $$$$=\int\frac{\sqrt{{tanx}}}{\mathrm{2}\frac{{sinx}}{{cosx}}×{cos}^{\mathrm{2}} {x}}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\:{sec}^{\mathrm{2}} {x}\sqrt{{tanx}}}{{tanx}}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{sec}^{\mathrm{2}} {x}}{\sqrt{{tanx}}}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\left({tanx}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {sec}^{\mathrm{2}} {xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\left({tanx}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\frac{\mathrm{1}}{\mathrm{2}}}\:+{c}=\sqrt{{tanx}}\:+{c} \\ $$

Commented by maxmathsup by imad last updated on 19/Apr/19

thanks sir

$${thanks}\:{sir} \\ $$

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