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Question Number 122587 by bramlexs22 last updated on 18/Nov/20

$$\underset{1}{\stackrel{16}{\int }}\arctan \sqrt{\sqrt{x}-1}dx$$$$\underset{0}{\stackrel{\pi /2}{\int }}\sin \left(2x\right)\arctan \left(\sin x\right)dx$$

Answered by MJS_new last updated on 18/Nov/20

$$\int \sin 2x\arctan \sin xdx=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=-\frac{1}{2}\cos 2x\arctan \sin x+\frac{1}{2}\int \frac{\cos x\cos 2x}{1+{\sin }^{2}x}dx=$$$$$$$$\frac{1}{2}\int \frac{\cos x\cos 2x}{1+{\sin }^{2}x}dx=$$$$[t=\sin x\to dx=\frac{dt}{\cos x}]$$$$=-\frac{1}{2}\int \frac{2t^2-1}{t^2+1}dt=-\int dt+\frac{3}{2}\int \frac{dt}{t^2+1}=$$$$=-t+\frac{3}{2}\arctan t=-\sin x+\frac{3}{2}\arctan \sin x$$$$$$$$=-\sin x+\frac{3-\cos 2x}{2}\arctan \sin x+C$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{\pi }{2}-1$$

Answered by liberty last updated on 18/Nov/20

$$\left(1\right)\underset{1}{\stackrel{16}{\int }}\arctan \sqrt{\sqrt{\mathrm{x}}-1}\mathrm{dx}$$$$\mathrm{letting}\sqrt{\sqrt{\mathrm{x}}-1}=\mathrm{♭}\to \{\begin{array}{l}\mathrm{♭}=0\\ \mathrm{♭}=\sqrt{3}\end{array}$$$$\Rightarrow \sqrt{\mathrm{x}}={\mathrm{♭}}^2+1\Rightarrow \mathrm{x}={({\mathrm{♭}}^2+1)}^2$$$$\Rightarrow \mathrm{dx}=4\mathrm{♭}({\mathrm{♭}}^2+1)\mathrm{d}\mathrm{♭}$$$$\mathrm{R}=\underset{0}{\stackrel{\sqrt{3}}{\int }}(4{\mathrm{♭}}^3+4\mathrm{♭})\arctan \mathrm{♭}\mathrm{d}\mathrm{♭}$$$$[\mathrm{D}.\mathrm{I}\mathrm{method}]$$$$\mathrm{R}={\left[({\mathrm{♭}}^4+2{\mathrm{♭}}^2)\arctan \mathrm{♭}\right]}_0^{\sqrt{3}}-\underset{0}{\stackrel{\sqrt{3}}{\int }}\left[\frac{{\mathrm{♭}}^4+2{\mathrm{♭}}^2}{1+{\mathrm{♭}}^2}\right]\mathrm{d}\mathrm{♭}$$$$\mathrm{R}=5\pi -\underset{0}{\stackrel{\sqrt{3}}{\int }}\frac{{(1+{\mathrm{♭}}^2)}^2-1}{1+{\mathrm{♭}}^2}\mathrm{d}\mathrm{♭}$$$$\mathrm{R}=5\pi -\underset{0}{\stackrel{\sqrt{3}}{\int }}(1+{\mathrm{♭}}^2)\mathrm{d}\mathrm{♭}+\underset{0}{\stackrel{\sqrt{3}}{\int }}\frac{\mathrm{d}\mathrm{♭}}{1+{\mathrm{♭}}^2}$$$${\mathrm{R}=5\pi -{(\mathrm{♭}+\frac{1}{3}{\mathrm{♭}}^3)}_0^{\sqrt{3}}+\arctan \left(\mathrm{♭}\right)]}_0^{\sqrt{3}}$$$$\mathrm{R}=5\pi -2\sqrt{3}+\frac{\pi }{3}=\frac{16\pi }{3}-2\sqrt{3}.▴$$

Answered by MJS_new last updated on 18/Nov/20

$$\int \arctan \sqrt{\sqrt{x}-1}dx=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=x\arctan \sqrt{\sqrt{x}-1}-\frac{1}{4}\int \frac{dx}{\sqrt{\sqrt{x}-1}}=$$$$$$$$-\frac{1}{4}\int \frac{dx}{\sqrt{\sqrt{x}-1}}=$$$$[t=\sqrt{\sqrt{x}-1}\to dx=4\sqrt{x}\sqrt{\sqrt{x}-1}dt]$$$$=-\int t^2+1dt=-\frac{t^3}{3}-t=$$$$=-\frac{(2+\sqrt{x})\sqrt{\sqrt{x}-1}}{3}$$$$$$$$=x\arctan \sqrt{\sqrt{x}-1}-\frac{(2+\sqrt{x})\sqrt{\sqrt{x}-1}}{3}+C$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{16\pi }{3}-2\sqrt{3}$$

Commented by liberty last updated on 18/Nov/20

$$\mathrm{why}\mathrm{my}\mathrm{answer}\mathrm{not}\mathrm{same}\mathrm{with}\mathrm{your}$$$$\mathrm{answer}\mathrm{sir}?\mathrm{what}\mathrm{wrong}?$$

Commented by MJS_new last updated on 18/Nov/20

$$\mathrm{I}\mathrm{made}\mathrm{a}\mathrm{mistake},{\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{correct}\mathrm{it}$$

Answered by Dwaipayan Shikari last updated on 18/Nov/20

$${\int }_1^{16}{tan}^{-1}\left(\sqrt{\sqrt{x}-1}\right)dxx=t^2\Rightarrow 1=2t\frac{dt}{dx}$$$${\int }_1^{4}2t{tan}^{-1}\left(\sqrt{t-1}\right)dtt-1=u^2\Rightarrow 1=2u\frac{du}{dt}$$$$4{\int }_0^{\sqrt{3}}u(u^2+1){tan}^{-1}udu$$$$\int u^3{tan}^{-1}udu=\frac{u^4}{4}{tan}^{-1}u-\frac{u^3}{12}+\frac{u}{4}-\frac{1}{4}{tan}^{-1}u$$$$\int {utan}^{-1}udu=\frac{u^2}{2}{tan}^{-1}u-\frac{u}{2}+\frac{1}{2}{tan}^{-1}u$$$$Soitbecomes$$$$4[\frac{9}{4}.\frac{\pi }{3}-\frac{3\sqrt{3}}{12}+\frac{\sqrt{3}}{4}-\frac{\pi }{12}]+4[\frac{3}{2}.\frac{\pi }{3}-\frac{\sqrt{3}}{2}+\frac{\pi }{6}]$$$$5\pi +\frac{\pi }{3}-2\sqrt{3}=\frac{16\pi }{3}-2\sqrt{3}$$

Answered by Bird last updated on 18/Nov/20

$$I={\int }_1^{16}srctan\left(\sqrt{\sqrt{x}-1}\right)dx$$$$changrment\sqrt{\sqrt{x}-1}=tgive$$$$\sqrt{x}-1=t^2\Rightarrow \sqrt{x}=t^2+1\Rightarrow$$$$x={(t^2+1)}^2\Rightarrow$$$$I={\int }_0^{\sqrt{3}}arctan\left(t\right)4t(t^2+1)dt$$$$=4{\int }_0^{\sqrt{3}}(t^3+t)arcctantdt$$$$=4{\left((\frac{t^4}{4}+\frac{t^2}{2})arctsnt\right]}_0^{\sqrt{3}}$$$$-{\int }_0^{\sqrt{3}}(\frac{t^4}{4}+\frac{t^2}{2})\frac{dt}{1+t^2}\}$$$$=4\left\{(\frac{9}{4}+\frac{3}{2})\frac{\pi }{3}\right\}-{\int }_0^{\sqrt{3}}\frac{t^4+2t^2}{t^2+1}dt$$$$=(9+6)\frac{\pi }{3}-{\int }_0^{\sqrt{3}}\frac{t^4+2t^2}{t^2+1}dt$$$$=5\pi -H$$$$H={\int }_0^{\sqrt{3}}\frac{t^2(t^2+1)+t^2}{t^2+1}dt$$$$={\int }_0^{\sqrt{3}}{t}^{2}dt+{\int }_0^{\sqrt{3}}\frac{t^2+1-1}{t^2+1}dt$$$$={\left[\frac{t^3}{3}\right]}_0^{\sqrt{3}}+\sqrt{3}-{\left[arctan\left(t\right)\right]}_0^{\sqrt{3}}$$$$=\sqrt{3}+\sqrt{3}-\frac{\pi }{3}=2\sqrt{3}-\frac{\pi }{3}\Rightarrow$$$$I=5\pi -2\sqrt{3}+\frac{\pi }{3}=\frac{16\pi }{3}-2\sqrt{3}$$$$$$

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