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Question Number 165328 by mnjuly1970 last updated on 30/Jan/22

$$$$$$provethat$$$$$$$$\mathcal{N}ice\mathcal{I}ntegral$$$$\mathit{\varphi }={\int }_0^1\frac{{tan}^{-1}\left(x^{\frac{3}{2}}\right)}{x^2}dx=\frac{\pi +\sqrt{3}ln(7+4\sqrt{3})}{4}◼m.n$$$$---------$$

Answered by Lordose last updated on 30/Jan/22

$$\mathrm{\Phi }={\int }_0^1\frac{{\tan }^{-1}\left({\mathrm{x}}^{\frac{3}{2}}\right)}{{\mathrm{x}}^2}\mathrm{dx}\stackrel{\mathbf{IBP}}{=}-\frac{{\tan }^{-1}\left({\mathrm{x}}^{\frac{3}{2}}\right)}{\mathrm{x}}{\mid }_0^1+\frac{3}{2}{\int }_0^1\frac{{\mathrm{x}}^{\frac{1}{2}-1}}{1+{\mathrm{x}}^3}\mathrm{dx}$$$$\mathrm{\Phi }=-\frac{\mathit{\pi }}{4}+\frac{3}{2}{\int }_0^1\frac{1}{\sqrt{\mathrm{x}}(1+{\mathrm{x}}^3)}\mathrm{dx}$$$$\mathrm{\Phi }=-\frac{\mathit{\pi }}{4}+3{\int }_0^1\frac{1}{(1+{\mathrm{x}}^6)}\mathrm{dx}$$$$\mathrm{\Phi }=-\frac{\mathit{\pi }}{4}+3\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}-1}{\int }_0^1{\mathrm{x}}^{6\mathrm{n}-6}\mathrm{dx}$$$$\mathrm{\Phi }=-\frac{\mathit{\pi }}{4}+3\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}}{6\mathrm{n}-5}$$$$\mathrm{\Phi }=-\frac{\mathit{\pi }}{4}+3(\frac{\mathit{\pi }}{6}+\frac{{\coth }^{-1}\left(\sqrt{3}\right)}{\sqrt{3}})$$$$\mathrm{\Phi }=\frac{\mathit{\pi }}{4}+\sqrt{3}{\coth }^{-1}\left(\sqrt{3}\right)$$

Answered by mnjuly1970 last updated on 30/Jan/22

$$---solution---$$$$\mathit{\varphi }\stackrel{i.b.p}{=}{\left[\frac{-1}{x}{tan}^{-1}\left(x^{\frac{3}{2}}\right)\right]}_0^1+\frac{3}{2}{\int }_0^1\frac{1}{\sqrt{x}.(1+x^3)}dx$$$$=-\frac{\pi }{4}+\frac{3}{2}\mathrm{\Omega }where\mathrm{\Omega }={\int }_0^1\frac{dx}{\sqrt{x}(1+x^3)}$$$$\mathrm{\Omega }\stackrel{\sqrt{x}=t}{=}{\int }_0^1\frac{2}{1+x^6}dx={\int }_0^1\frac{x^4+1-(x^4-1)}{1+x^6}dx$$$$={\int }_0^1\frac{(x^4-x^2+1)+x^2}{1+x^6}dx+{\int }_0^1\frac{1-x^2}{1-x^2+x^4}dx$$$$=\frac{\pi }{4}+{\int }_0^1\frac{3x^2}{1+{\left(x^3\right)}^2}dx-{\int }_0^1\frac{1-x^{-2}}{{(x+x^{-1})}^2-3}dx$$$$=\frac{\pi }{4}+\frac{\pi }{4}+{\int }_2^{\mathrm{\infty }}\frac{dx}{x^2-3}$$$$=\frac{\pi }{2}+{\int }_2^{\mathrm{\infty }}\frac{dx}{(x-\sqrt{3})(x+\sqrt{3})}$$$$=\frac{\pi }{2}+\frac{1}{2\sqrt{3}}\left\{{\left[ln\left(\frac{x-\sqrt{3}}{x+\sqrt{3}}\right)\right]}_2^{\mathrm{\infty }}\right\}$$$$=\frac{\pi }{2}+\frac{1}{2\sqrt{3}}ln\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)$$$$=\frac{\pi }{2}+\frac{1}{2\sqrt{3}}ln(7+4\sqrt{3})$$$$\therefore \mathit{\varphi }=\frac{\pi }{4}+\frac{\sqrt{3}}{4}ln(7+4\sqrt{3})◼m.n$$$$$$

Answered by Ar Brandon last updated on 30/Jan/22

$$\mathit{\varphi }={\int }_0^1\frac{{\tan }^{-1}\left(x^{\frac{3}{2}}\right)}{x^2}dx=-{\left[\frac{1}{x}{\tan }^{-1}\left(x^{\frac{3}{2}}\right)\right]}_0^1+\frac{3}{2}{\int }_0^1\frac{x^{-\frac{1}{2}}}{1+x^3}dx$$$$=-\frac{\pi }{4}+3{\int }_0^1\frac{dt}{1+t^6}=-\frac{\pi }{4}+3{\int }_0^1\frac{dt}{(t^2+1)(t^4-t^2+1)}$$$$=-\frac{\pi }{4}+{\int }_0^1(\frac{1}{t^2+1}-\frac{t^2-2}{t^4-t^2+1})dt=-\frac{\pi }{4}+\frac{\pi }{4}-{\int }_0^1\frac{t^2-2}{t^4-t^2+1}dt$$$$=-\frac{1}{2}{\int }_0^1\frac{3(t^2-1)-(t^2+1)}{t^4-t^2+1}dt=\frac{1}{2}{\int }_0^1\frac{t^2+1}{t^4-t^2+1}dt-\frac{3}{2}{\int }_0^1\frac{t^2-1}{t^4-t^2+1}dt$$$$=\frac{1}{2}{\int }_0^1\frac{1+\frac{1}{t^2}}{{(t-\frac{1}{t})}^2+1}dt-\frac{3}{2}{\int }_0^1\frac{1-\frac{1}{t^2}}{{(t+\frac{1}{t})}^2-3}dt$$$$=\frac{1}{2}{\left[\arctan \left(\frac{t^2-1}{t}\right)\right]}_0^1+\frac{\sqrt{3}}{4}{[\ln \mid \frac{t^2+\sqrt{3}t+1}{t^2-\sqrt{3}t+1}\mid ]}_0^1$$$$=\frac{1}{2}\left(\frac{\pi }{2}\right)+\frac{\sqrt{3}}{4}\ln \mid \frac{2+\sqrt{3}}{2-\sqrt{3}}\mid =\frac{\pi }{4}+\frac{\sqrt{3}}{4}\ln {(2+\sqrt{3})}^2$$$$=\frac{\pi }{4}+\frac{\sqrt{3}}{4}\ln (7+4\sqrt{3})$$

Commented by mnjuly1970 last updated on 31/Jan/22

$$thanksalotsirbrandon$$$$$$

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