If int frac{sin x}{3+4 cos ^2 x} ,dx = A tan | vedclass

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(D) Let $I = \int \frac{\sin x}{3+4 \cos ^2 x} \,dx$.Substitute $\cos x = t$, then $-\sin x \,dx = dt$, which implies $\sin x \,dx = -dt$.Substituting

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$\frac{\sqrt{3}}{2}$

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(D) Let $I = \int \frac{\sin x}{3+4 \cos ^2 x} \,dx$.Substitute $\cos x = t$, then $-\sin x \,dx = dt$, which implies $\sin x \,dx = -dt$.Substituting these into the integral, we get $I = \int \frac{-dt}{3+4 t^2} = -\int \frac{dt}{(\sqrt{3})^2 + (2t)^2}$.Using the standard integral formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$, we have:$I = -\frac{1}{\sqrt{3}} \cdot \frac{1}{2} 	an^{-1}(\frac{2t}{\sqrt{3}}) + C = -\frac{1}{2\sqrt{3}} 	an^{-1}(\frac{2 \cos x}{\sqrt{3}}) + C$.Comparing this with $A 	an^{-1}(B \cos x) + C$, we identify $A = -\frac{1}{2\sqrt{3}}$ and $B = \frac{2}{\sqrt{3}}$.Therefore, $A + B = -\frac{1}{2\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{-1 + 4}{2\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.

If $\int \frac{\sin x}{3+4 \cos ^2 x} \,dx = A \tan ^{-1}(B \cos x) + C$, (where $C$ is a constant of integration), then the value of $A+B$ is

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