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

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(D) Let $I = \int \frac{\cos 	heta}{5+7 \sin 	heta-2 \cos ^2 	heta} d 	heta$. Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get: $I =

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

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(D) Let $I = \int \frac{\cos 	heta}{5+7 \sin 	heta-2 \cos ^2 	heta} d 	heta$. Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get: $I = \int \frac{\cos 	heta}{5+7 \sin 	heta-2(1-\sin^2 	heta)} d 	heta = \int \frac{\cos 	heta}{2 \sin^2 	heta+7 \sin 	heta+3} d 	heta$. Factoring the denominator: $2 \sin^2 	heta+7 \sin 	heta+3 = (2 \sin 	heta+1)(\sin 	heta+3)$. Let $\sin 	heta = t$,then $\cos 	heta d 	heta = dt$. $I = \int \frac{dt}{(2t+1)(t+3)}$. Using partial fractions: $\frac{1}{(2t+1)(t+3)} = \frac{A'}{2t+1} + \frac{B'}{t+3} \Rightarrow 1 = A'(t+3) + B'(2t+1)$. For $t = -3$,$1 = B'(-5) \Rightarrow B' = -\frac{1}{5}$. For $t = -\frac{1}{2}$,$1 = A'(\frac{5}{2}) \Rightarrow A' = \frac{2}{5}$. $I = \int (\frac{2/5}{2t+1} - \frac{1/5}{t+3}) dt = \frac{1}{5} \ln |2t+1| - \frac{1}{5} \ln |t+3| + c = \frac{1}{5} \ln |\frac{2 \sin 	heta+1}{\sin 	heta+3}| + c$. Comparing with $A \ln |f(	heta)| + c$,we have $A = \frac{1}{5}$ and $f(	heta) = \frac{2 \sin 	heta+1}{\sin 	heta+3}$. Thus,$\frac{f(	heta)}{A} = \frac{(2 \sin 	heta+1)/(\sin 	heta+3)}{1/5} = \frac{5(2 \sin 	heta+1)}{\sin 	heta+3}$.

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

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