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Question Number 122927 by bemath last updated on 21/Nov/20

Commented by liberty last updated on 21/Nov/20

$$\mu \left(x\right)=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$$$$[let\sqrt{x}=\sin z\Rightarrow dx=\sin 2zdz]$$$$\mu \left(x\right)=\int \sqrt{\frac{1-\sin z}{1+\sin z}}\left(\sin 2zdz\right)$$$$\mu \left(x\right)=\int \frac{\sqrt{{(1-\sin z)}^2}}{\cos z}\left(2\sin z\cos z\right)dz$$$$\mu \left(x\right)=\int 2\sin z(1-\sin z)dz$$$$\mu \left(x\right)=\int (2\sin z-2(\frac{1}{2}-\cos 2z))dz$$$$\mu \left(x\right)=-2\cos z-z+\sin 2z+c$$$$\mu \left(x\right)=2\cos z(\sin z-1)-z+c$$$$\mu \left(x\right)=2\sqrt{1-x}(\sqrt{x}-1)-{\sin }^{-1}\sqrt{x}+c$$

Answered by MJS_new last updated on 21/Nov/20

$$\mathrm{we}\mathrm{had}\mathrm{this}\mathrm{a}\mathrm{few}\mathrm{days}\mathrm{ago}\mathrm{but}\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{find}$$$$\mathrm{the}\mathrm{old}\mathrm{post}$$$$\mathrm{simply}\mathrm{let}t=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\iff x=\frac{{(t^2-1)}^2}{{(t^2+1)}^2}\wedge t^2-1⩽0$$$$\to dx=-2\sqrt{x}\sqrt{1-\sqrt{x}}{(1+\sqrt{x})}^{3/2}dt$$$$\mathrm{the}\mathrm{integral}\mathrm{now}\mathrm{becomes}$$$$8\int \frac{t^2(t^2-1)}{{(t^2+1)}^3}dt$$$$\mathrm{and}\mathrm{this}\mathrm{is}‘‘{\mathrm{easy}}^{″}\mathrm{to}\mathrm{solve}.\mathrm{I}\mathrm{get}$$$$2\arctan t-\frac{2t(3t^2+1)}{{(t^2+1)}^2}=$$$$=(\sqrt{x}-2)\sqrt{1-x}+2\arctan \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C=$$$$=(\sqrt{x}-2)\sqrt{1-x}+\arccos \sqrt{x}+C$$

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