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Question Number 114472 by mnjuly1970 last updated on 19/Sep/20

$$...calculus...$$$$evaluate::$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{tan\left(2x\right)}{\sqrt{{sin}^4\left(x\right)+4{cos}^2\left(x\right)}-\sqrt{{cos}^4\left(x\right)+4{sin}^2\left(x\right)}}dx=???$$$$$$$$...m.n.july.1970....$$$$$$

Commented by Dwaipayan Shikari last updated on 19/Sep/20

$${\int }_0^{\frac{\pi }{2}}\frac{tan2x}{{sin}^{4}x+4{cos}^{2}x-{cos}^{4}x-4{sin}^{2}x}(\sqrt{{sin}^{4}x+4{cos}^{2}x}+\sqrt{{cos}^{4}x+4{sin}^{2}x})dx$$$${\int }_0^{\frac{\pi }{2}}\frac{tan2x}{3cos2x}.(\sqrt{{sin}^{4}x-4{sin}^{2}x+4}+\sqrt{{cos}^{4}x-4{cos}^{2}x+4})$$$$\frac{1}{3}{\int }_0^{\frac{\pi }{2}}\frac{tan2x}{cos2x}.({sin}^{2}x-2+{cos}^{2}x-2)$$$$\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{-2sin2x}{{cos}^{2}2x}dx({sin}^{4}x-{cos}^{4}x={sin}^{2}x-{cos}^{2}x=-cos2x)$$$$\frac{1}{2}{\int }_1^{-1}\frac{dt}{t^2}$$$$-\frac{1}{2}{\int }_{-1}^1\frac{dt}{t^2}={\left[\frac{1}{2t}\right]}_{-1}^1\to Diverges$$

Commented by mnjuly1970 last updated on 19/Sep/20

$$peacebeuponyoumr$$$$payan.thankyouso$$$$much...$$

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