int frac{sin x , dx}{a^2 + b^2 cos^2 x} = | vedclass

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(C) Let $I = \int \frac{\sin x \, dx}{a^2 + b^2 \cos^2 x}$.Substitute $t = b \cos x$. Then $dt = -b \sin x \, dx$,which implies $\sin x \, dx = -\frac

Vedclass · Accepted Answer

$\frac{1}{ab} \cot^{-1} \left( \frac{b \cos x}{a} \right) + c$

Vedclass · Answer

(C) Let $I = \int \frac{\sin x \, dx}{a^2 + b^2 \cos^2 x}$.Substitute $t = b \cos x$. Then $dt = -b \sin x \, dx$,which implies $\sin x \, dx = -\frac{dt}{b}$.The integral becomes $I = \int \frac{-dt/b}{a^2 + t^2} = -\frac{1}{b} \int \frac{dt}{a^2 + t^2}$.Using the standard integral formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1} \left( \frac{x}{a} ight) + c$,we get $I = -\frac{1}{b} \cdot \frac{1}{a} 	an^{-1} \left( \frac{t}{a} ight) + c = -\frac{1}{ab} 	an^{-1} \left( \frac{b \cos x}{a} ight) + c$.Using the identity $	an^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$,we have $-	an^{-1}(x) = \cot^{-1}(x) - \frac{\pi}{2}$.Thus,$I = \frac{1}{ab} \cot^{-1} \left( \frac{b \cos x}{a} ight) + C$,where $C = c - \frac{\pi}{2ab}$ is a constant.Therefore,the correct option is $C$.

$\int \frac{\sin x \, dx}{a^2 + b^2 \cos^2 x} = $

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