int_0^{pi /4} {frac{{4sin 2theta ,dtheta }}{{ | vedclass

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(C) Let $I = \int_0^{\pi /4} {\frac{{4\sin 2	heta \,d	heta }}{{{{\sin }^4}	heta + {{\cos }^4}	heta }}} = \int_0^{\pi /4} {\frac{{8\sin 	heta \cos

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$\pi $

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(C) Let $I = \int_0^{\pi /4} {\frac{{4\sin 2	heta \,d	heta }}{{{{\sin }^4}	heta + {{\cos }^4}	heta }}} = \int_0^{\pi /4} {\frac{{8\sin 	heta \cos 	heta \,d	heta }}{{{{\sin }^4}	heta + {{\cos }^4}	heta }}} $Dividing the numerator and denominator by $\cos^4 	heta$,we get:$I = \int_0^{\pi /4} {\frac{{8	an 	heta \sec^2 	heta \,d	heta }}{{{	an^4 	heta + 1}}} } $Let $	an^2 	heta = t$. Then $2 	an 	heta \sec^2 	heta \,d	heta = dt$.When $	heta = 0$,$t = 0$. When $	heta = \pi/4$,$t = 1$.Substituting these into the integral:$I = 4 \int_0^1 {\frac{{dt}}{{{t^2} + 1}}} $$I = 4 [	an^{-1} t]_0^1 = 4(\frac{\pi}{4} - 0) = \pi $.

$\int_0^{\pi /4} {\frac{{4\sin 2\theta \,d\theta }}{{{{\sin }^4}\theta + {{\cos }^4}\theta }}} = $

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