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

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Question

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

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$\frac{1}{b(a + b \cos x)} + c$

Vedclass · Answer

(B) Let $I = \int \frac{\sin x}{(a + b \cos x)^2} \, dx$. Substitute $t = a + b \cos x$. Then,$dt = -b \sin x \, dx$,which implies $\sin x \, dx = -\frac{dt}{b}$. Substituting these into the integral: $I = \int \frac{1}{t^2} \left(-\frac{dt}{b}\right) = -\frac{1}{b} \int t^{-2} \, dt$. Integrating $t^{-2}$ gives $-t^{-1} = -\frac{1}{t}$. Thus,$I = -\frac{1}{b} \left(-\frac{1}{t}\right) + c = \frac{1}{bt} + c$. Substituting back $t = a + b \cos x$,we get $I = \frac{1}{b(a + b \cos x)} + c$.

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

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