The value of int frac{1}{3 sin x-cos x+3} d x | vedclass

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(C) Let $I = \int \frac{dx}{3 \sin x - \cos x + 3}$.Using the substitution $	an \frac{x}{2} = t$,we have $dx = \frac{2 dt}{1+t^2}$,$\sin x = \frac{2t

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$\log \left(\frac{2 	an \frac{x}{2}+1}{	an \frac{x}{2}+1}ight)+C$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{3 \sin x - \cos x + 3}$.Using the substitution $	an \frac{x}{2} = t$,we have $dx = \frac{2 dt}{1+t^2}$,$\sin x = \frac{2t}{1+t^2}$,and $\cos x = \frac{1-t^2}{1+t^2}$.Substituting these into the integral:$I = \int \frac{1}{3(\frac{2t}{1+t^2}) - (\frac{1-t^2}{1+t^2}) + 3} \cdot \frac{2 dt}{1+t^2}$$I = \int \frac{2 dt}{6t - 1 + t^2 + 3(1+t^2)} = \int \frac{2 dt}{4t^2 + 6t + 2} = \int \frac{dt}{2t^2 + 3t + 1}$.Factor the denominator: $2t^2 + 3t + 1 = (2t+1)(t+1)$.Using partial fractions: $\frac{1}{(2t+1)(t+1)} = \frac{A}{2t+1} + \frac{B}{t+1}$.Solving for $A$ and $B$,we get $A = -2$ and $B = 1$.$I = \int (\frac{1}{t+1} - \frac{2}{2t+1}) dt = \log|t+1| - \log|2t+1| + C = \log \left| \frac{t+1}{2t+1} ight| + C$.Wait,checking the partial fraction again: $\frac{1}{(2t+1)(t+1)} = \frac{2}{2t+1} - \frac{1}{t+1}$.$I = \int (\frac{2}{2t+1} - \frac{1}{t+1}) dt = \log|2t+1| - \log|t+1| + C = \log \left| \frac{2t+1}{t+1} ight| + C$.Substituting $t = 	an \frac{x}{2}$,we get $I = \log \left( \frac{2 	an \frac{x}{2} + 1}{	an \frac{x}{2} + 1} ight) + C$.

The value of $\int \frac{1}{3 \sin x-\cos x+3} d x$ is

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