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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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Verified Answer
The correct answer is:
$\log \left|\frac{1+\cos x}{\cos x}\right|$
$\text { Let } I=\int \frac{\sin x}{\cos x(1+\cos x)} d x$
Put $\cos x=t \Rightarrow-\sin x d x=d t$
$\begin{aligned}
\Rightarrow \quad I & =\int \frac{-d t}{t(1+t)} \\
& =-\int\left[\frac{1}{t}-\frac{1}{(1+t)}\right] d t \\
& =-[\log t-\log (1+t)]+c \\
& =\log \left(\frac{t+1}{t}\right)+c
\end{aligned}$
$\begin{aligned}
& \text { But } I=\int(x)+c \\
& \therefore \quad \log \left(\frac{\cos x+1}{\cos x}\right)+c=f(x)+c \\
& \Rightarrow \quad f(x)=\log \left(\frac{1+\cos x}{\cos x}\right)
\end{aligned}$
Put $\cos x=t \Rightarrow-\sin x d x=d t$
$\begin{aligned}
\Rightarrow \quad I & =\int \frac{-d t}{t(1+t)} \\
& =-\int\left[\frac{1}{t}-\frac{1}{(1+t)}\right] d t \\
& =-[\log t-\log (1+t)]+c \\
& =\log \left(\frac{t+1}{t}\right)+c
\end{aligned}$
$\begin{aligned}
& \text { But } I=\int(x)+c \\
& \therefore \quad \log \left(\frac{\cos x+1}{\cos x}\right)+c=f(x)+c \\
& \Rightarrow \quad f(x)=\log \left(\frac{1+\cos x}{\cos x}\right)
\end{aligned}$
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