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Question Number 99173 by bemath last updated on 19/Jun/20

Answered by MJS last updated on 19/Jun/20

$$\mathrm{this}\mathrm{is}\mathrm{very}\mathrm{complicated}$$$$\int \sqrt{\frac{1+\cot x}{\csc x+\cot x}}dx=\int \sqrt{\frac{\sin x+\cos x}{1+\cos x}}dx=$$$$[t=\tan \frac{x}{2}\to dx={2\cos }^2\frac{x}{2}dt]$$$$=\sqrt{2}\int \frac{\sqrt{-t^2+2t+1}}{t^2+1}dt=$$$$[u=\frac{\sqrt{2}}{2}(t-1)\to dt=\sqrt{2}du]$$$$=\sqrt{2}\int \frac{\sqrt{1-u^2}}{u^2+\sqrt{2}u+1}du=$$$$[v={\cos }^{-1}u\to du=-\sqrt{1-u^2}dv]$$$$=-\sqrt{2}\int \frac{{\sin }^{2}v}{{\cos }^{2}v+\sqrt{2}\cos v+1}dv=$$$$[w=\tan \frac{v}{2}\to dv={2\cos }^2\frac{v}{2}dw]$$$$=-8(1+\sqrt{2})\int \frac{w^2}{(w^2+1)(w^4+3+2\sqrt{2})}dw$$$$2\sqrt{2}\int \frac{dw}{w^2+1}+\sqrt{-2+2\sqrt{2}}(\int \frac{w-\sqrt{7+5\sqrt{2}}}{w^2-\sqrt{2+2\sqrt{2}}w+1+\sqrt{2}}dw-\int \frac{w+\sqrt{7+5\sqrt{2}}}{w^2+\sqrt{2+2\sqrt{2}}w+1+\sqrt{2}}dw)$$$$\mathrm{now}\mathrm{just}\mathrm{use}\mathrm{formula}$$

Commented by bemath last updated on 19/Jun/20

$$\mathrm{yes}..\mathrm{sir}\mathrm{very}\mathrm{hard}...$$

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