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Question Number 160792 by cortano last updated on 06/Dec/21

$$\int \frac{\sec \mathrm{x}}{\sqrt{1+2\sec \mathrm{x}}}\sqrt{\frac{\mathrm{cosec}\mathrm{x}-\cot \mathrm{x}}{\mathrm{cosec}\mathrm{x}+\cot \mathrm{x}}}\mathrm{dx}=?$$

Answered by chhaythean last updated on 06/Dec/21

$$=\int \frac{\frac{1}{\mathrm{cosx}}}{\sqrt{\frac{\mathrm{cosx}+2}{\mathrm{cosx}}}}\times \sqrt{\frac{\frac{1-\mathrm{cosx}}{\mathrm{sinx}}}{\frac{1+\mathrm{cosx}}{\mathrm{sinx}}}}\mathrm{dx}$$$$=\int \frac{\frac{1}{\mathrm{cosx}}}{\sqrt{\frac{\mathrm{cosx}+2}{\mathrm{cosx}}}}\times \sqrt{\frac{1-\mathrm{cosx}}{1+\mathrm{cosx}}}\mathrm{dx}$$$$=\int \frac{\mathrm{sinx}}{\sqrt{{\cos }^2\mathrm{x}+2\mathrm{cosx}}}\times \frac{1}{1+\mathrm{cosx}}\mathrm{dx}$$$$=\int \frac{\mathrm{sinx}}{\sqrt{{(\mathrm{cosx}+1)}^2-1}(1+\mathrm{cosx})}\mathrm{dx}$$$$\mathrm{let}\mathrm{u}=1+\mathrm{cosx}\Rightarrow \mathrm{du}=-\mathrm{sinxdx}$$$$=-\int \frac{\mathrm{du}}{\sqrt{{\mathrm{u}}^2-1}\times \mathrm{u}}$$$$\mathrm{let}\mathrm{u}=\mathrm{secy}\Rightarrow \mathrm{du}=\mathrm{secytanydy}$$$$=-\int \frac{\mathrm{secytany}}{\mathrm{tanysecy}}\mathrm{dy}=-\mathrm{y}+\mathrm{c}$$$$=-\mathrm{arcsec}\left(\mathrm{u}\right)+\mathrm{c}$$$$=-\mathrm{arcsec}(1+\mathrm{cosx})+\mathrm{c}$$$$\mathrm{So}\overline{)\begin{array}{c}\int \frac{\mathrm{secx}}{\sqrt{1+2\mathrm{secx}}}\times \sqrt{\frac{\mathrm{cosecx}-\mathrm{cotx}}{\mathrm{cosecx}+\mathrm{cotx}}}\mathrm{dx}=-\mathrm{arcsec}(1+\mathrm{cosx})+\mathrm{c}\end{array}}$$

Answered by MJS_new last updated on 06/Dec/21

$$\mathrm{let}c=\cos x\to dx=-\frac{dc}{\sqrt{1-c^2}}$$$$\mathrm{now}\mathrm{we}\mathrm{have}$$$$-\int \frac{dc}{(c+1)\sqrt{c}\sqrt{c+2}}=$$$$[t=\frac{\sqrt{c+2}}{\sqrt{c}}\to dc=-\sqrt{c^3}\sqrt{c+2}dt]$$$$=2\int \frac{dt}{t^2+1}=2\arctan t=2\arctan \frac{\sqrt{c+2}}{\sqrt{c}}=$$$$=2\arctan \frac{\sqrt{2+\cos x}}{\sqrt{\cos x}}+C$$

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