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Question Number 211797 by Spillover last updated on 21/Sep/24

Answered by Ghisom last updated on 21/Sep/24

$$\int \frac{\sqrt{\cot x}-\sqrt{\tan x}}{1+3\sin 2x}dx=$$$$[t=\sqrt{\tan x}]$$$$=-2\int \frac{t^2-1}{t^4+6t^2+1}dt=$$$$[u=\frac{2t}{t^2+1}]$$$$=\int \frac{du}{u^2+1}=\arctan u=\arctan \frac{2t}{t^2+1}=$$$$=\arctan \frac{2\sqrt{\tan x}}{1+\tan x}=$$$$=\arctan \frac{2\sqrt{\cos x\sin x}}{\cos x+\sin x}+C$$

Commented by TonyCWX08 last updated on 22/Sep/24

$$Ibegyourpardon,butIthinkthisiswrong.$$

Commented by Ghisom last updated on 22/Sep/24

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{convinced}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{right}$$$$\mathrm{but}\mathrm{feel}\mathrm{free}\mathrm{to}\mathrm{prove}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}$$

Commented by TonyCWX08 last updated on 22/Sep/24

$$Itshouldbe1-t^{2}afterthesubstitution$$

Commented by Frix last updated on 22/Sep/24

$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{correct}:$$$$2\int \frac{1-t^2}{t^4+6t^2+1}dt=-2\int \frac{t^2-1}{t^4+6t^2+1}dt$$

Answered by TonyCWX08 last updated on 22/Sep/24

$$I=\int \frac{\sqrt{cot\left(x\right)}-\sqrt{tan\left(x\right)}}{1+3sin\left(2x\right)}dx$$$$=\int \frac{1}{\sqrt{tan\left(x\right)}}\left(\frac{1-tan\left(x\right)}{1+6sin\left(x\right)cos\left(x\right)}\right)dx$$$$=\int \frac{1}{\sqrt{tan\left(x\right)}}\left(\frac{1-tan\left(x\right)}{{sec}^2\left(x\right)+6tan\left(x\right)}\right)\left({sec}^2\left(x\right)\right)dx$$$$=\int \frac{1}{\sqrt{tan\left(x\right)}}\left(\frac{1-tan\left(x\right)}{{tan}^2\left(x\right)+6tan\left(x\right)+1}\right)\left({sec}^2\left(x\right)\right)dx$$$$lett=\sqrt{tan\left(x\right)}$$$$dt=\frac{{sec}^2\left(x\right)dx}{2\sqrt{tan\left(x\right)}}\Rightarrow 2tdt={sec}^2\left(x\right)dx$$$$=\int \frac{1}{t}\left(\frac{1-t^2}{t^4+6t^2+1}\right)\left(2tdt\right)$$$$=2\int \left(\frac{1-t^2}{t^4+6t^2+1}\right)dt$$$$=2\int \left(\frac{1-\frac{1}{t^2}}{t^2+6+\frac{1}{t^2}}\right)dt$$$$=2\int \left(\frac{1-\frac{1}{t^2}}{{(t+\frac{1}{t})}^2+4}\right)dt$$$$Letv=t+\frac{1}{t}$$$$dv=(1-\frac{1}{t^2})dt$$$$=2\int \frac{dv}{v^2+4}$$$$=2\int \left(\frac{dv}{v^2+4}\right)$$$$=2\left(\frac{{\tan }^{-1}\left(\frac{v}{2}\right)}{2}\right)+c$$$$={\tan }^{-1}\left(\frac{1}{2}(t+\frac{1}{t})\right)+c$$$$={\tan }^{-1}\left(\frac{1}{2}(\sqrt{tan\left(x\right)}+\frac{1}{\sqrt{tan\left(x\right)}})\right)+c$$$$={\tan }^{-1}\left(\frac{1}{2}(\sqrt{tan\left(x\right)}+\sqrt{cot\left(x\right)})\right)+c$$$$={\tan }^{-1}\left(\frac{\sqrt{tan\left(x\right)}+\sqrt{cot\left(x\right)}}{2}\right)+C$$

Commented by TonyCWX08 last updated on 22/Sep/24

$$Hereismysolution.$$$$InspiredbyDr.Ghisom$$

Commented by Frix last updated on 22/Sep/24

$$2\int \frac{1-t^2}{t^4+6t^2+1}dt=2\int \frac{\frac{1}{t^2}-1}{t^2+6+\frac{1}{t^2}}dt=\stackrel{!}{-}2\int \frac{1-\frac{1}{t^2}}{t^2+6+\frac{1}{t^2}}dt$$

Commented by Spillover last updated on 23/Sep/24

$$correct.thanks$$

Answered by Spillover last updated on 24/Sep/24

Answered by Spillover last updated on 24/Sep/24

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