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Question Number 207565 by mathzup last updated on 18/May/24

$$find{\int }_0^{\frac{\pi }{2}}\frac{x^2}{{tan}^{2}x}dx$$

Answered by sniper237 last updated on 19/May/24

$${\int }_0^{\pi /2}\frac{x^2(1-{sin}^{2}x)}{{sin}^{2}x}dx={\int }_0^{\pi /2}\frac{x^2}{{sin}^{2}x}dx-\frac{{\pi }^3}{24}$$$$\stackrel{PI}{=}{[-\frac{x^2}{tanx}]}_{\to 0}+2{\int }_0^{\pi /2}\frac{x}{tanx}dx-\frac{{\pi }^3}{24}$$$$\stackrel{PI}{=}2{\left[xln\left(sinx\right)\right]}_{\to 0}-2{\int }_0^{\pi /2}ln\left(sinx\right)dx-\frac{{\pi }^3}{24}$$$$\stackrel{t=sinx}{=}-2{\int }_0^1\frac{lnt}{\sqrt{1-t^2}}dt-\frac{{\pi }^3}{24}$$$$=-2{\partial }_a{\left({\int }_0^1\frac{t^a}{\sqrt{1-t^2}}dt\right)}_0-\frac{{\pi }^3}{24}$$$$\stackrel{u=t^2}{=}-{\partial }_a\left({\int }_0^1{u}^{\frac{a-1}{2}}{(1-u)}^{\frac{1}{2}-1}du\right)-\frac{{\pi }^3}{24}$$$$=-{\partial }_a{\left(\frac{\mathrm{\Gamma }\left(\frac{a+1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\frac{a}{2}+1)}\right)}_0-\frac{{\pi }^3}{24}$$$$=-\frac{1}{2}\frac{{\mathrm{\Gamma }}^{\prime }\left(1/2\right)\mathrm{\Gamma }\left(1/2\right)-{\mathrm{\Gamma }}^{\prime }\left(1\right)\mathrm{\Gamma }{\left(1/2\right)}^2}{1^2}-\frac{{\pi }^3}{24}$$$$=\frac{3\pi \gamma -6\sqrt{\pi }{\mathrm{\Gamma }}^{\prime }\left(1/2\right)-{\pi }^3}{24}$$

Answered by Berbere last updated on 19/May/24

$$x\to \frac{\pi }{2}-x$$$$={\int }_0^{\frac{\pi }{2}}{(\frac{\pi }{2}-x)}^2\tan {}^2\left(x\right)dx={\int }_0^{\frac{\pi }{2}}(1+\tan {}^2\left(x\right)){(\frac{\pi }{2}-x)}^{2}dx$$$$-{\int }_0^{\frac{\pi }{2}}{(\frac{\pi }{2}-x)}^2=A-B$$$$B=-\frac{1}{3}{\left[{(\frac{\pi }{2}-x)}^3\right]}_0^{\frac{\pi }{2}}=\frac{{\pi }^3}{24}$$$$A={\left[\tan \left(x\right){(\frac{\pi }{2}-x)}^2\right]}_0^{\frac{\pi }{2}}-2{\int }_0^{\frac{\pi }{2}}\tan \left(x\right)(\frac{\pi }{2}-x)$$$$=-2{\int }_0^{\frac{\pi }{2}}\tan \left(x\right)(\frac{\pi }{2}-x)dx={\left[2ln\left(cos\left(x\right)\right)(\frac{\pi }{2}-x)\right]}_0^{\frac{\pi }{2}}$$$$-2{\int }_0^{\frac{\pi }{2}}ln\left(cos\left(x\right)\right)dx=-2.-\frac{\pi }{2}ln\left(2\right)=\pi ln\left(2\right)$$$$\pi ln\left(2\right)-\frac{{\pi }^3}{24}$$

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