Question Number 114472 by mnjuly1970 last updated on 19/Sep/20
$$\:\:\:\:\:\:\:\:\:...\:\:{calculus}... \\ $$$${evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{tan}\left(\mathrm{2}{x}\right)}{\:\sqrt{{sin}^{\mathrm{4}} \left({x}\right)+\mathrm{4}{cos}^{\mathrm{2}} \left({x}\right)}−\sqrt{{cos}^{\mathrm{4}} \left({x}\right)+\mathrm{4}{sin}^{\mathrm{2}} \left({x}\right)\:}}\:{dx}=\:??? \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:...{m}.{n}.{july}.\mathrm{1970}.... \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 19/Sep/20
![∫_0 ^(π/2) ((tan2x)/(sin^4 x+4cos^2 x−cos^4 x−4sin^2 x))((√(sin^4 x+4cos^2 x )) +(√(cos^4 x+4sin^2 x)) )dx ∫_0 ^(π/2) ((tan2x)/(3cos2x)).((√(sin^4 x−4sin^2 x+4)) +(√(cos^4 x−4cos^2 x+4))) (1/3)∫_0 ^(π/2) ((tan2x)/(cos2x)).(sin^2 x−2+cos^2 x−2) (1/2)∫_0 ^(π/2) ((−2sin2x)/(cos^2 2x))dx (sin^4 x−cos^4 x=sin^2 x−cos^2 x=−cos2x) (1/2)∫_1 ^(−1) (dt/t^2 ) −(1/2)∫_(−1) ^1 (dt/t^2 )=[(1/(2t))]_(−1) ^1 →Diverges](Q114473.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tan}\mathrm{2}{x}}{{sin}^{\mathrm{4}} {x}+\mathrm{4}{cos}^{\mathrm{2}} {x}−{cos}^{\mathrm{4}} {x}−\mathrm{4}{sin}^{\mathrm{2}} {x}}\left(\sqrt{{sin}^{\mathrm{4}} {x}+\mathrm{4}{cos}^{\mathrm{2}} {x}\:}\:+\sqrt{{cos}^{\mathrm{4}} {x}+\mathrm{4}{sin}^{\mathrm{2}} {x}}\:\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tan}\mathrm{2}{x}}{\mathrm{3}{cos}\mathrm{2}{x}}.\left(\sqrt{{sin}^{\mathrm{4}} {x}−\mathrm{4}{sin}^{\mathrm{2}} {x}+\mathrm{4}}\:+\sqrt{{cos}^{\mathrm{4}} {x}−\mathrm{4}{cos}^{\mathrm{2}} {x}+\mathrm{4}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tan}\mathrm{2}{x}}{{cos}\mathrm{2}{x}}.\left({sin}^{\mathrm{2}} {x}−\mathrm{2}+{cos}^{\mathrm{2}} {x}−\mathrm{2}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{−\mathrm{2}{sin}\mathrm{2}{x}}{{cos}^{\mathrm{2}} \mathrm{2}{x}}{dx}\:\:\:\:\:\:\left({sin}^{\mathrm{4}} {x}−{cos}^{\mathrm{4}} {x}={sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}=−{cos}\mathrm{2}{x}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{−\mathrm{1}} \frac{{dt}}{{t}^{\mathrm{2}} } \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{dt}}{{t}^{\mathrm{2}} }=\left[\frac{\mathrm{1}}{\mathrm{2}{t}}\right]_{−\mathrm{1}} ^{\mathrm{1}} \rightarrow{Diverges} \\ $$
Commented by mnjuly1970 last updated on 19/Sep/20
$${peace}\:{be}\:{upon}\:{you}\:{mr} \\ $$$${payan}.{thank}\:{you}\:{so} \\ $$$${much}... \\ $$