int frac{sin x}{3+4 cos ^2 x} d x | vedclass

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Question

(A) Let $I = \int \frac{\sin x}{3+4 \cos ^2 x} d x$. Substitute $u = \cos x$,then $du = -\sin x d x$,or $\sin x d x = -du$. Substituting these into th

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{\sin x}{3+4 \cos ^2 x} d x$. Substitute $u = \cos x$,then $du = -\sin x d x$,or $\sin x d x = -du$. Substituting these into the integral gives $I = \int \frac{-du}{3+4u^2} = -\int \frac{du}{3+(2u)^2}$. Factor out $3$ from the denominator: $I = -\frac{1}{3} \int \frac{du}{1 + (\frac{2u}{\sqrt{3}})^2}$. Let $t = \frac{2u}{\sqrt{3}}$,then $dt = \frac{2}{\sqrt{3}} du$,so $du = \frac{\sqrt{3}}{2} dt$. Substituting $t$ into the integral: $I = -\frac{1}{3} \int \frac{\frac{\sqrt{3}}{2} dt}{1+t^2} = -\frac{\sqrt{3}}{6} \int \frac{dt}{1+t^2}$. Since $\frac{\sqrt{3}}{6} = \frac{1}{2\sqrt{3}}$,we have $I = -\frac{1}{2\sqrt{3}} 	an^{-1}(t) + C$. Substituting back $t = \frac{2 \cos x}{\sqrt{3}}$,we get $I = -\frac{1}{2\sqrt{3}} 	an^{-1}\left(\frac{2 \cos x}{\sqrt{3}}ight) + C$.

$\int \frac{\sin x}{3+4 \cos ^2 x} d x$

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