int frac{2-sin x}{2 cos x+3} d x= | vedclass

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(B) Let $I = \int \frac{2-\sin x}{2 \cos x+3} dx$.We can split the integral into two parts: $I = \int \frac{2}{2 \cos x+3} dx - \int \frac{\sin x}{2 \

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$\frac{4}{\sqrt{5}} 	an ^{-1}\left(\frac{1}{\sqrt{5}} 	an \frac{x}{2}ight)+\log \sqrt{2 \cos x+3}+c$

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(B) Let $I = \int \frac{2-\sin x}{2 \cos x+3} dx$.We can split the integral into two parts: $I = \int \frac{2}{2 \cos x+3} dx - \int \frac{\sin x}{2 \cos x+3} dx = I_1 + I_2$.For $I_1 = \int \frac{2}{2 \cos x+3} dx$,use the half-angle substitution $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)} = \frac{1-t^2}{1+t^2}$ where $t = 	an(x/2)$ and $dx = \frac{2 dt}{1+t^2}$.$I_1 = \int \frac{2}{2(\frac{1-t^2}{1+t^2})+3} \cdot \frac{2 dt}{1+t^2} = \int \frac{4}{2-2t^2+3+3t^2} dt = \int \frac{4}{t^2+5} dt$.$I_1 = \frac{4}{\sqrt{5}} 	an^{-1}(\frac{t}{\sqrt{5}}) = \frac{4}{\sqrt{5}} 	an^{-1}(\frac{1}{\sqrt{5}} 	an \frac{x}{2}) + C_1$.For $I_2 = \int \frac{-\sin x}{2 \cos x+3} dx$,let $u = 2 \cos x + 3$,then $du = -2 \sin x dx$,so $-\sin x dx = \frac{1}{2} du$.$I_2 = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \log|2 \cos x + 3| = \log \sqrt{2 \cos x + 3} + C_2$.Combining these,$I = \frac{4}{\sqrt{5}} 	an^{-1}(\frac{1}{\sqrt{5}} 	an \frac{x}{2}) + \log \sqrt{2 \cos x + 3} + C$.

$\int \frac{2-\sin x}{2 \cos x+3} d x=$

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