If int frac{cos theta}{5+7 sin theta-2 cos ^{ | vedclass

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(D) Given integral: $I = \int \frac{\cos 	heta d 	heta}{5+7 \sin 	heta-2 \cos ^{2} 	heta}$Substituting $\cos^2 	heta = 1 - \sin^2 	heta$:$I = \i

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$\frac{5(2 \sin 	heta+1)}{\sin 	heta+3}$

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(D) Given integral: $I = \int \frac{\cos 	heta d 	heta}{5+7 \sin 	heta-2 \cos ^{2} 	heta}$Substituting $\cos^2 	heta = 1 - \sin^2 	heta$:$I = \int \frac{\cos 	heta d 	heta}{5+7 \sin 	heta-2(1-\sin^2 	heta)} = \int \frac{\cos 	heta d 	heta}{2 \sin^2 	heta+7 \sin 	heta+3}$Let $\sin 	heta = t$,then $\cos 	heta d 	heta = dt$:$I = \int \frac{dt}{2t^2+7t+3} = \int \frac{dt}{(2t+1)(t+3)}$Using partial fractions:$\frac{1}{(2t+1)(t+3)} = \frac{1}{5} \left( \frac{2}{2t+1} - \frac{1}{t+3} ight)$Integrating:$I = \frac{1}{5} \int \left( \frac{2}{2t+1} - \frac{1}{t+3} ight) dt = \frac{1}{5} \ln \left| \frac{2t+1}{t+3} ight| + C$Substituting $t = \sin 	heta$ back:$I = \frac{1}{5} \ln \left| \frac{2 \sin 	heta+1}{\sin 	heta+3} ight| + C$Comparing with $A \log _{e}|B(	heta)|+C$,we get $A = \frac{1}{5}$ and $B(	heta) = \frac{2 \sin 	heta+1}{\sin 	heta+3}$.Therefore,$\frac{B(	heta)}{A} = \frac{\frac{2 \sin 	heta+1}{\sin 	heta+3}}{\frac{1}{5}} = \frac{5(2 \sin 	heta+1)}{\sin 	heta+3}$.

If $\int \frac{\cos \theta}{5+7 \sin \theta-2 \cos ^{2} \theta} d \theta=A \log _{e}|B(\theta)|+C$ where $C$ is a constant of integration,then $\frac{ B (\theta)}{ A }$ can be

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