Integrate the function frac{sin x}{1+cos x}. | vedclass

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Question

(A) Let $I = \int \frac{\sin x}{1+\cos x} dx$. Substitute $t = 1 + \cos x$. Differentiating with respect to $x$,we get $dt = -\sin x dx$,which implies

Vedclass · Accepted Answer

$-\log |1+\cos x| + C$

Vedclass · Answer

(A) Let $I = \int \frac{\sin x}{1+\cos x} dx$. Substitute $t = 1 + \cos x$. Differentiating with respect to $x$,we get $dt = -\sin x dx$,which implies $\sin x dx = -dt$. Substituting these into the integral,we get $I = \int -\frac{dt}{t}$. Integrating,we obtain $I = -\log |t| + C$. Substituting back $t = 1 + \cos x$,we get $I = -\log |1 + \cos x| + C$,where $C$ is an arbitrary constant.

Integrate the function $\frac{\sin x}{1+\cos x}$.

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