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Question Number 40717 by ajeetyadav4370 last updated on 26/Jul/18

$$\int \sqrt{tanx/sinx.cosxdx}$$

Commented by math khazana by abdo last updated on 26/Jul/18

$$letI=\int \sqrt{\frac{tanx}{sinxcosx}}dx$$$$I=\int \sqrt{\frac{sinx}{sinx{cos}^{2}x}}dx$$$$=\int \frac{dx}{cosx}changementtan\left(\frac{x}{2}\right)=tgive$$$$I=\int \frac{1}{\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}=\int \frac{2dt}{1-t^2}$$$$=\int \{\frac{1}{1+t}+\frac{1}{1-t}\}dt=ln\mid \frac{1+t}{1-t}\mid +c$$$$=ln\mid \frac{1+tan\left(\frac{x}{2}\right)}{1-tan\left(\frac{x}{2}\right)}\mid +c=ln\mid tan(\frac{x}{2}+\frac{\pi }{4})\mid +c.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

$$\int \sqrt{\frac{tanx}{sinxcosx}}dx$$$$\int secxdx$$$$\int \frac{dx}{cosx}$$$$\int \frac{1+{tan}^2\frac{x}{2}}{1-{tan}^2\frac{x}{2}}dxlett=tan\frac{x}{2}dt={sec}^2\frac{x}{2}.\frac{1}{2}.dx$$$$\int \frac{1+t^2}{1-t^2}\times \frac{2dt}{1+t^2}$$$$2\int \frac{dt}{1-t^2}$$$$\int \frac{1+t+1-t}{(1+t)(1-t)}dt$$$$\int \frac{dt}{1-t}+\int \frac{dt}{1+t}$$$$ln(1+t)-ln(1-t)+c$$$$ln\left(\frac{1+t}{1-t}\right)+c$$$$ln\left(\frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}\right)+c$$$$ln\left(\frac{1+2tan\frac{x}{2}+{tan}^2\frac{x}{2}}{1-{tan}^2\frac{x}{2}}\right)+c$$$$ln\left(\frac{1+\frac{2tan\frac{x}{2}}{1+{tan}^2\frac{x}{2}}}{\frac{1-{tan}^2\frac{x}{2}}{1+{tan}^2\frac{x}{2}}}\right)=ln\left(\frac{1+sinx}{cosx}\right)=ln(secx+tanx)$$$$$$

Commented by $@ty@m last updated on 26/Jul/18

$$\int \sec xdx$$$$\int \sec x.\frac{(\sec x+\tan x)}{(\sec x+\tan x)}dx$$$$=\ln (\sec x+\tan x)+C$$

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