int frac{1}{3 cos x - 4 sin x + 5} dx = | vedclass

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(C) We use the substitution $	an \frac{x}{2} = t$,which implies $dx = \frac{2 dt}{1 + t^2}$,$\cos x = \frac{1 - t^2}{1 + t^2}$,and $\sin x = \frac{2t

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$\frac{1}{2 - 	an \frac{x}{2}} + c$

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(C) We use the substitution $	an \frac{x}{2} = t$,which implies $dx = \frac{2 dt}{1 + t^2}$,$\cos x = \frac{1 - t^2}{1 + t^2}$,and $\sin x = \frac{2t}{1 + t^2}$.Substituting these into the integral:$\int \frac{1}{3(\frac{1 - t^2}{1 + t^2}) - 4(\frac{2t}{1 + t^2}) + 5} \cdot \frac{2 dt}{1 + t^2}$$= \int \frac{2 dt}{3(1 - t^2) - 8t + 5(1 + t^2)}$$= \int \frac{2 dt}{3 - 3t^2 - 8t + 5 + 5t^2}$$= \int \frac{2 dt}{2t^2 - 8t + 8} = \int \frac{dt}{t^2 - 4t + 4}$$= \int \frac{dt}{(t - 2)^2} = -(t - 2)^{-1} + c$$= \frac{-1}{t - 2} + c = \frac{1}{2 - t} + c$Substituting back $t = 	an \frac{x}{2}$,we get $\frac{1}{2 - 	an \frac{x}{2}} + c$.

$\int \frac{1}{3 \cos x - 4 \sin x + 5} dx = $

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