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Question Number 132121 by abdullahquwatan last updated on 11/Feb/21

$${\int }_2^8\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}\mathrm{dx}$$

Commented by Dwaipayan Shikari last updated on 11/Feb/21

$$3$$

Commented by abdullahquwatan last updated on 11/Feb/21

$$thx$$$$answerialreadyknow,theway{don}^{\prime }tknow$$

Answered by EDWIN88 last updated on 11/Feb/21

$$\mathrm{let}\sqrt{\mathrm{x}}=\sqrt{10}\sin \mathrm{u}\mathrm{or}\mathrm{x}=10\sin {}^2\mathrm{u}\Rightarrow \mathrm{dx}=20\sin \mathrm{u}\cos \mathrm{u}\mathrm{du}$$$$\mathrm{I}=\int \frac{\sqrt{10}\sin \mathrm{u}}{\sqrt{10-10\sin {}^2\mathrm{u}}+\sqrt{10}\sin \mathrm{u}}\left(20\sin \mathrm{u}\cos \mathrm{u}\mathrm{du}\right)$$$$\mathrm{I}=\int \frac{20\sqrt{10}{\sin }^2\mathrm{u}\cos \mathrm{u}}{\sqrt{10}(\cos \mathrm{u}+\sin \mathrm{u})}\mathrm{du}$$$$\mathrm{I}=10\int \frac{\sin 2\mathrm{u}\cos \mathrm{u}}{\cos 2\mathrm{u}}(\cos \mathrm{u}-\sin \mathrm{u})\mathrm{du}$$$$\mathrm{I}=10\int \tan 2\mathrm{u}(\frac{1}{2}+\frac{1}{2}\cos 2\mathrm{u})\mathrm{du}-\frac{1}{2}\int \tan 2\mathrm{u}\sin 2\mathrm{u}\mathrm{du}$$$$=-\frac{5}{2}\int \frac{\mathrm{d}\left(\cos 2\mathrm{u}\right)}{\cos 2\mathrm{u}}+5\int \sin 2\mathrm{u}\mathrm{du}-\frac{1}{2}\int \frac{1-\cos {}^{2}2\mathrm{u}}{\cos 2\mathrm{u}}\mathrm{du}$$$$=-\frac{5}{2}\ln \mid \cos 2\mathrm{u}\mid -\frac{5}{2}\cos 2\mathrm{u}-\frac{1}{4}\ln \mid \sec 2\mathrm{u}+\tan 2\mathrm{u}\mid +\frac{1}{4}\sin 2\mathrm{u}$$$$\mathrm{I}={[-\frac{5}{2}\ln \mid \frac{5-\mathrm{x}}{5}\mid -\frac{1}{2}(5-\mathrm{x})-\frac{1}{4}\ln \mid \frac{5+\sqrt{10\mathrm{x}-{\mathrm{x}}^2}}{5-\mathrm{x}}\mid +\frac{\sqrt{10\mathrm{x}-{\mathrm{x}}^2}}{20}]}_2^8$$$$\mathrm{I}=-\frac{5}{2}(\ln \frac{3}{5}-\ln \frac{3}{5})-\frac{1}{2}(-3-3)-\frac{1}{4}(\ln 3-\ln 3)+\frac{4-4}{20}$$$$\mathrm{I}=-\frac{1}{2}(-6)=3$$

Commented by abdullahquwatan last updated on 11/Feb/21

$$thx$$

Answered by Dwaipayan Shikari last updated on 11/Feb/21

$${\int }_2^8\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}dx={\int }_2^8\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}dx=I$$$$2I={\int }_2^8\frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{x}+\sqrt{10-x}}dx$$$$2I=(8-2)\Rightarrow I=3$$

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