If int frac{sin x}{cos x(1+cos x)} d x=f(x)+c | vedclass

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(A) Let $I = \int \frac{\sin x}{\cos x(1+\cos x)} dx$. Substitute $\cos x = t$,then $-\sin x dx = dt$,which implies $\sin x dx = -dt$. Substituting th

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$\log \left|\frac{1+\cos x}{\cos x}\right|$

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(A) Let $I = \int \frac{\sin x}{\cos x(1+\cos x)} dx$. Substitute $\cos x = t$,then $-\sin x dx = dt$,which implies $\sin x dx = -dt$. Substituting these into the integral,we get: $I = \int \frac{-dt}{t(1+t)}$. Using partial fractions,$\frac{1}{t(1+t)} = \frac{1}{t} - \frac{1}{1+t}$. Therefore,$I = -\int \left( \frac{1}{t} - \frac{1}{1+t} \right) dt$. $I = -(\log |t| - \log |1+t|) + c$. $I = \log |1+t| - \log |t| + c = \log \left| \frac{1+t}{t} \right| + c$. Substituting $t = \cos x$ back,we get $I = \log \left| \frac{1+\cos x}{\cos x} \right| + c$. Since $I = f(x) + c$,we have $f(x) = \log \left| \frac{1+\cos x}{\cos x} \right|$.

If $\int \frac{\sin x}{\cos x(1+\cos x)} d x=f(x)+c$,then $f(x)$ is equal to

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