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

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Question

(C) 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 t

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) 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 $I = \int \frac{-dt}{3 + 4t^2}$.$I = -\frac{1}{4} \int \frac{dt}{t^2 + (\frac{\sqrt{3}}{2})^2}$.Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$,we get:$I = -\frac{1}{4} \cdot \frac{1}{\frac{\sqrt{3}}{2}} 	an^{-1} \left( \frac{t}{\frac{\sqrt{3}}{2}} ight) + c$.$I = -\frac{1}{4} \cdot \frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{2t}{\sqrt{3}} ight) + c$.$I = -\frac{1}{2\sqrt{3}} 	an^{-1} \left( \frac{2\cos x}{\sqrt{3}} ight) + c$.

$\int \frac{\sin x \, dx}{3 + 4\cos^2 x} = $

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