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Question Number 76298 by benjo last updated on 26/Dec/19

$$\int \sqrt{\mathrm{tanx}+\mathrm{cotxdx}}$$

Commented by benjo last updated on 26/Dec/19

Commented by john santuy last updated on 26/Dec/19

$$usingWeirstrrasmetode.$$$$lett=tan\left(\frac{x}{2}\right)$$

Commented by benjo last updated on 26/Dec/19

$$\mathrm{please}\mathrm{sir}\mathrm{your}\mathrm{write}\mathrm{step}\mathrm{by}\mathrm{step}$$

Commented by mathmax by abdo last updated on 26/Dec/19

$$letI=\int \sqrt{tanx+\frac{1}{tanx}}dx\Rightarrow I=\int \sqrt{\frac{sinx}{cosx}+\frac{cosx}{sinx}}dx$$$$=\int \sqrt{\frac{1}{cosxsinx}}dx=\int \frac{1}{\sqrt{\frac{1}{2}sin\left(2x\right)}}dx=\int \frac{\sqrt{2}}{\sqrt{sin\left(2x\right)}}dx$$$$vhangement\sqrt{sin\left(2x\right)}=tgiveI=\int \frac{\sqrt{2}}{t}\frac{tdt}{\sqrt{1-t^4}}because$$$$sin\left(2x\right)=t^2\Rightarrow 2x=arcsin\left(t^2\right)\Rightarrow 2dx=\frac{2tdt}{\sqrt{1-t^4}}$$$$\Rightarrow I=\int \frac{\sqrt{2}dt}{\sqrt{1-t^4}}andthisintegralisnotresolublebyelementary$$$$functions..$$

Answered by john santuy last updated on 26/Dec/19

$$I=\int secx\sqrt{cotx}dx$$$$letcotx=u^2\to -{csc}^{2}xdx=2udu$$$$dx=-\frac{2udu}{\sqrt{1+u^4}}$$$$I=\int \frac{\sqrt{1+u^4}}{u^2}\times u^2\times \frac{2u}{\sqrt{1+u^4}}du$$$$=-2\int udu=-u^2+C$$$$hence-cotx+C◼$$

Commented by benjo last updated on 26/Dec/19

$$\mathrm{waw}\mathrm{thanks}\mathrm{sir}$$

Commented by MJS last updated on 26/Dec/19

$$\mathrm{sorry}\mathrm{but}{\mathrm{that}}^{\prime }\mathrm{s}\mathrm{wrong}$$$$\int \sqrt{\tan x+\cot x}dx=\int \frac{dx}{\sqrt{\sin x\cos x}}=$$$$=\sqrt{2}\int \frac{dx}{\sqrt{\sin 2x}}$$$$\mathrm{and}\mathrm{this}\mathrm{cannot}\mathrm{be}\mathrm{solved}\mathrm{using}\mathrm{elementary}$$$$\mathrm{calculus}$$

Commented by MJS last updated on 26/Dec/19

$$\mathrm{but}\frac{d}{dx}[-\cot x]={\csc }^{2}x\ne \sec x\sqrt{\cot x}$$

Commented by john santu last updated on 26/Dec/19

$$\mathrm{tanx}+\frac{1}{\mathrm{tanx}}=\frac{{\tan }^2\mathrm{x}+1}{\mathrm{tanx}}=\frac{{\sec }^2\mathrm{x}}{\mathrm{tanx}}$$$$\sqrt{\frac{{\sec }^2\mathrm{x}}{\mathrm{tanx}}}=\mathrm{secx}\sqrt{\mathrm{cotx}}\mathrm{sir}?$$

Commented by john santu last updated on 26/Dec/19

$$ooyessir.i^{\prime }mwrongindx=-\frac{2u}{{\left(\sqrt{1+u^2}\right)}^2}du$$

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