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Question Number 174341 by behi834171 last updated on 31/Jul/22

$$1.\underset{1}{\stackrel{2}{\int }}\sqrt{\mathit{x}+\sqrt{\mathit{x}-1}}\mathit{d}\mathit{x}=?$$$$2.\underset{0}{\stackrel{\frac{\mathit{\pi }}{2}}{\int }}\frac{2\mathit{s}\mathit{i}\mathit{n}\mathit{x}+4\mathit{c}\mathit{o}\mathit{s}\mathit{x}}{\mathit{s}\mathit{i}\mathit{n}\mathit{x}+\mathit{c}\mathit{o}\mathit{s}\mathit{x}}\mathit{d}\mathit{x}=?$$

Commented by mnjuly1970 last updated on 30/Jul/22

$$2.\left(3\pi /2\right)$$

Answered by MJS_new last updated on 30/Jul/22

$$1.$$$$\int \sqrt{x+\sqrt{x-1}}dx=$$$$[t=\sqrt{x-1}+\frac{1}{2}\to dx=(2t-1)dt]$$$$=\frac{1}{2}\int (2t-1)\sqrt{4t^2+3}dt=$$$$[u=\frac{2t+\sqrt{4t^2+3}}{\sqrt{3}}\to dt=\frac{\sqrt{4t^2+3}}{2u}du]$$$$=\int (\frac{3\sqrt{3}}{32}{u}^2-\frac{3}{16}u+\frac{3\sqrt{3}}{32}-\frac{3}{8}{u}^{-1}-\frac{3\sqrt{3}}{32}{u}^{-2}-\frac{3}{16}{u}^{-3}-\frac{3\sqrt{3}}{32}{u}^{-4})du$$$$\mathrm{the}\mathrm{rest}\mathrm{is}\mathrm{easy}$$

Commented by behi834171 last updated on 30/Jul/22

$$verynice!$$$$thankyousomuchdearmaster.$$

Answered by Frix last updated on 30/Jul/22

$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{2\sin x+4\cos x}{\sin x+\cos x}dx=$$$$=3\underset{0}{\stackrel{\pi /2}{\int }}dx-\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sin x-\cos x}{\sin x+\cos x}dx$$$$3\underset{0}{\stackrel{\pi /2}{\int }}dx=\frac{3\pi }{2}$$$$-\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sin x-\cos x}{\sin x+\cos x}dx=$$$$=\underset{0}{\stackrel{\pi /2}{\int }}\frac{d[\sin x+\cos x]}{\sin x+\cos x}=$$$$={[\ln \mid \sin x+\cos x\mid ]}_0^{\pi /2}=0$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{3\pi }{2}$$

Commented by behi834171 last updated on 30/Jul/22

$$thankyouverymuchsir.$$$$youransweristrue.$$

Answered by behi834171 last updated on 31/Jul/22

$$let:x-1=t^2\Rightarrow \mathit{d}\mathit{x}=2\mathit{t}\mathit{d}\mathit{t}$$$$\mathit{I}=\int 2\mathit{t}\sqrt{{\mathit{t}}^2+1+\mathit{t}}\mathit{d}\mathit{t}=\int (2\mathit{t}+1-1)\sqrt{{\mathit{t}}^2+\mathit{t}+1}\mathit{d}\mathit{t}=$$$$=\int (2\mathit{t}+1)\sqrt{{\mathit{t}}^2+\mathit{t}+1}\mathit{d}\mathit{t}-\int \sqrt{{\mathit{t}}^2+\mathit{t}+1}\mathit{d}\mathit{t}={\mathit{I}}_1+{\mathit{I}}_2$$$${\mathit{I}}_1=\frac{2}{3}{({\mathit{t}}^2+\mathit{t}+1)}^{\frac{3}{2}}+\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}.$$$${\mathit{I}}_2=\int \sqrt{{(\mathit{t}+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\mathit{d}\mathit{t}=$$$$=\frac{2\mathit{t}+1}{4}\sqrt{{\mathit{t}}^2+\mathit{t}+1}+\frac{3}{8}{\mathit{s}\mathit{i}\mathit{n}\mathit{h}}^{-1}\left(\frac{2\mathit{t}+1}{\sqrt{3}}\right)+\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}.$$$$\Rightarrow \mathit{I}=\frac{2}{3}{(\mathit{x}+\sqrt{\mathit{x}-1})}^{\frac{3}{2}}+\left(\frac{2\sqrt{\mathit{x}-1}+1}{4}\right).\sqrt{\mathit{x}+\sqrt{\mathit{x}-1}}+$$$$+$$$$\frac{3}{8}{\mathit{s}\mathit{i}\mathit{n}\mathit{h}}^{-1}\left(\frac{2\sqrt{\mathit{x}-1}+1}{\sqrt{3}}\right)+\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}.◼$$$$[I_1=2\sqrt{3}-\frac{2}{3}=\frac{6\sqrt{3}-2}{3}$$$$I_2=\frac{3\sqrt{3}-1}{4}+\frac{3}{10}=\frac{15\sqrt{3}+1}{20}$$$$\Rightarrow \mathbf{I}={\mathit{I}}_1+{\mathit{I}}_2=\frac{165\sqrt{3}-37}{60}]$$

Commented by peter frank last updated on 01/Aug/22

$$\mathrm{thanks}$$

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