The integral 80 int_0^{frac{pi}{4}} left( fra | vedclass

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(C) Let $I = 80 \int_0^{\frac{\pi}{4}} \frac{\sin 	heta + \cos 	heta}{9 + 16 \sin 2 	heta} d 	heta$. We know that $\sin 2 	heta = 1 - (1 - \sin 2

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$4 \log_e 3$

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(C) Let $I = 80 \int_0^{\frac{\pi}{4}} \frac{\sin 	heta + \cos 	heta}{9 + 16 \sin 2 	heta} d 	heta$. We know that $\sin 2 	heta = 1 - (1 - \sin 2 	heta) = 1 - (\sin 	heta - \cos 	heta)^2$. Substituting this into the integral: $I = 80 \int_0^{\frac{\pi}{4}} \frac{\sin 	heta + \cos 	heta}{9 + 16(1 - (\sin 	heta - \cos 	heta)^2)} d 	heta = 80 \int_0^{\frac{\pi}{4}} \frac{\sin 	heta + \cos 	heta}{25 - 16(\sin 	heta - \cos 	heta)^2} d 	heta$. Let $t = \sin 	heta - \cos 	heta$,then $dt = (\cos 	heta + \sin 	heta) d 	heta$. When $	heta = 0$,$t = -1$. When $	heta = \frac{\pi}{4}$,$t = 0$. $I = 80 \int_{-1}^0 \frac{dt}{25 - 16t^2} = 80 \int_{-1}^0 \frac{dt}{16(\frac{25}{16} - t^2)} = 5 \int_{-1}^0 \frac{dt}{(\frac{5}{4})^2 - t^2}$. Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln |\frac{a+x}{a-x}|$: $I = 5 \left[ \frac{1}{2(\frac{5}{4})} \ln \left| \frac{\frac{5}{4} + t}{\frac{5}{4} - t} ight| ight]_{-1}^0 = 5 \left[ \frac{2}{5} \ln \left| \frac{5+4t}{5-4t} ight| ight]_{-1}^0 = 2 [\ln(1) - \ln(\frac{1}{9})] = 2 [0 - \ln(3^{-2})] = 2 [2 \ln 3] = 4 \ln 3$.

The integral $80 \int_0^{\frac{\pi}{4}} \left( \frac{\sin \theta + \cos \theta}{9 + 16 \sin 2 \theta} \right) d \theta$ is equal to :

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