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Question Number 194963 by C2coder last updated on 20/Jul/23

Answered by witcher3 last updated on 21/Jul/23

$${\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\frac{2\mathrm{asin}\left(\mathrm{x}\right)}{1-{\mathrm{a}}^2{\sin }^2\left(\mathrm{x}\right)}\right)\mathrm{dx}..?$$$$$$$$$$$$$$

Commented by C2coder last updated on 22/Jul/23

$$ithoughtthesameanfwasable$$$$tosolveitbutithinkthis$$$$formiscorrectjustharder$$$$andmorecomplex$$

Answered by witcher3 last updated on 21/Jul/23

$${\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\frac{{2\mathrm{asin}}^2\left(\mathrm{x}\right)}{1-{\mathrm{a}}^2{\sin }^2\left(\mathrm{x}\right)}\right)\mathrm{dx}=\mathrm{f}\left(\mathrm{a}\right);\mathrm{f}\left(0\right)=0\mathrm{w}$$$${\mathrm{f}}^{\prime }\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{2}}\frac{{2\sin }^2\left(\mathrm{x}\right)(1+{\mathrm{a}}^2{\sin }^2\left(\mathrm{x}\right))}{{(1-{\mathrm{a}}^2{\sin }^2\left(\mathrm{x}\right))}^2+{\left({2\mathrm{asin}}^2\left(\mathrm{x}\right)\right)}^2}$$$$=\frac{1}{2}{\int }_0^{2\pi }\frac{{\sin }^2\left(\mathrm{x}\right)(1+{\mathrm{a}}^2{\sin }^2\left(\mathrm{x}\right))}{(-{\mathrm{a}}^2{\sin }^2\left(\mathrm{x}\right)+{2\mathrm{aisin}}^2\left(\mathrm{x}\right)+1)(-{\mathrm{a}}^2{\sin }^2\left(\mathrm{x}\right)-{2\mathrm{iasin}}^2\left(\mathrm{x}\right)+1)}$$$$$$$$=2{\int }_0^{\frac{\pi }{2}}\frac{{\sin }^2\left(\mathrm{x}\right)(1+{\mathrm{a}}^2{\sin }^2\left(\mathrm{x}\right))}{{(\mathrm{ia}+1)}^2{\sin }^2\left(\mathrm{x}\right)+{\cos }^2\left(\mathrm{x}\right))({(1-\mathrm{ia})}^2{\sin }^2\left(\mathrm{x}\right)+{\cos }^2\left(\mathrm{x}\right))}\mathrm{dx}$$$$$$$$=2{\int }_0^{\frac{\pi }{2}}\frac{{\mathrm{tg}}^2\left(\mathrm{x}\right)(1+(1+{\mathrm{a}}^2){\mathrm{tg}}^2\left(\mathrm{x}\right))}{(1+{(\mathrm{ia}+1)}^2{\mathrm{tg}}^2\left(\mathrm{x}\right))(1+{(1-\mathrm{ia})}^2{\mathrm{tg}}^2\left(\mathrm{x}\right))}$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{t}}^2(1+(1+{\mathrm{a}}^2){\mathrm{t}}^2)}{(1+{\mathrm{t}}^2)(1+{(1+\mathrm{ia})}^2{\mathrm{t}}^2)(1+{(1-\mathrm{ia})}^2{\mathrm{t}}^2)}{}_{=\mathrm{g}}\mathrm{dt}$$$$=2\mathrm{i}\pi \mathrm{Res}(\mathrm{z},\mathrm{g};\mathrm{im}\left(\mathrm{z}\right)>0)$$$$\mathrm{a}>0$$$$1+{(1+\mathrm{ia})}^2{\mathrm{t}}^2=0\Rightarrow \mathrm{t}=\frac{\mathrm{i}}{1+\mathrm{ia}}$$$$1+{\mathrm{t}}^2=0\Rightarrow \mathrm{t}=\mathrm{i}$$$$1+{(1-\mathrm{ia})}^2{\mathrm{t}}^2=0\Rightarrow \mathrm{t}=\frac{\mathrm{i}}{1-\mathrm{ia}}$$$$2\mathrm{i}\pi .(\frac{-(-{\mathrm{a}}^2)}{2\mathrm{i}(1-{(1-\mathrm{ia})}^2)(1-{(1+\mathrm{ia})}^2)}$$$$2\mathrm{i}\pi (\frac{-\frac{1}{{(1+\mathrm{ia})}^2}(1-(1+{\mathrm{a}}^2).{\left(\frac{1}{1+\mathrm{ia}}\right)}^2)}{(1-{\left(\frac{1}{1+\mathrm{ia}}\right)}^2)\left(2\mathrm{i}(1+\mathrm{ia})\right)(1-\frac{{(1-\mathrm{ia})}^2}{{(1+\mathrm{ia})}^2})}$$$$2\mathrm{i}\pi (\frac{-\frac{1}{{(1-\mathrm{ia})}^2}(1-(1+{\mathrm{a}}^2).\frac{1}{{(1-\mathrm{ia})}^2})}{(1-\frac{1}{{(1-\mathrm{ia})}^2})(1-{\left(\frac{1+\mathrm{ia}}{1-\mathrm{ia}}\right)}^2)\left(2\mathrm{i}(1-\mathrm{ia})\right)}$$$$2\mathrm{i}\pi \left(\frac{{\mathrm{a}}^2}{2\mathrm{i}({\mathrm{a}}^2+2\mathrm{ia})({\mathrm{a}}^2-2\mathrm{ia})}\right),\frac{\pi }{{\mathrm{a}}^2+4},\frac{\pi }{2}{\tan }^{-1}\left(\mathrm{a}\right)$$$$+2\mathrm{i}\pi (\frac{({2\mathrm{a}}^2-2\mathrm{ia})}{(-{\mathrm{a}}^2+2\mathrm{ia})(2\mathrm{i}-2\mathrm{a})\left(4\mathrm{ia}\right)}$$$$+2\mathrm{i}\pi (\frac{({2\mathrm{a}}^2+2\mathrm{ia})}{(-4\mathrm{ia})(-{\mathrm{a}}^2-2\mathrm{ia})(2\mathrm{i}+2\mathrm{a})}$$$$2\mathrm{i}\pi (\frac{2\mathrm{a}-2\mathrm{i}}{(2\mathrm{i}-\mathrm{a})(2\mathrm{i}-2\mathrm{a})\left(4\mathrm{ia}\right)}-\frac{(2\mathrm{a}+2\mathrm{i})}{4\mathrm{ia}(-\mathrm{a}-2\mathrm{i})(2\mathrm{i}+2\mathrm{a})}$$$$\frac{\pi }{2\mathrm{a}}(\frac{(2\mathrm{a}-2\mathrm{i})(-\mathrm{a}-2\mathrm{i})(2\mathrm{i}+2\mathrm{a})-(2\mathrm{a}+2\mathrm{i})(2\mathrm{i}-\mathrm{a})(2\mathrm{i}-2\mathrm{a})}{({\mathrm{a}}^2+4)(-4-{4\mathrm{a}}^2)}$$$$\frac{\pi }{2\mathrm{a}}(\frac{(-{2\mathrm{a}}^2-2\mathrm{ia}-4)(2\mathrm{a}+2\mathrm{i})-(-{2\mathrm{a}}^2+2\mathrm{ia}-4)(2\mathrm{i}-2\mathrm{a})}{({\mathrm{a}}^2+4)(-4-{4\mathrm{a}}^2)}$$$$\frac{\pi }{2\mathrm{a}}(\frac{-{4\mathrm{a}}^3+{\mathrm{a}}^2(-8\mathrm{i})+\mathrm{a}(-4)-8\mathrm{i}-({4\mathrm{a}}^3+{\mathrm{a}}^2(-8\mathrm{i})+\mathrm{a}\left(4\right)-8\mathrm{i})}{({\mathrm{a}}^2+4)(-4-{4\mathrm{a}}^2)}$$$$=\frac{\pi }{{\mathrm{a}}^2+4}+\frac{\pi }{2\mathrm{a}}\left(\frac{-{8\mathrm{a}}^3-8\mathrm{a}}{({\mathrm{a}}^2+4)(-4-{4\mathrm{a}}^2)}\right)$$$$=\frac{2\pi }{{\mathrm{a}}^2+4}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{a}}{\mathrm{f}}^{\prime }\left(\mathrm{t}\right)\mathrm{dt},\mid \mathrm{f}\left(0\right)=0$$$$={\int }_0^{\mathrm{a}}\frac{2\pi }{{\mathrm{t}}^2+4}\mathrm{dt}=\pi {\int }_0^{\mathrm{a}}\frac{\mathrm{d}\left(\frac{\mathrm{t}}{2}\right)}{(1+\left(\frac{\mathrm{t}}{2}\right)}=\pi {\tan }^{-1}\left(\frac{\mathrm{a}}{2}\right)$$$${\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\frac{{2\mathrm{asin}}^2\left(\mathrm{x}\right)}{1-{\mathrm{a}}^2{\sin }^2\left(\mathrm{x}\right)}\right)\mathrm{dx}=\pi {\tan }^{-1}\left(\frac{\mathrm{a}}{2}\right)$$

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