$\begin{aligned}= & \int \frac{\sin (n x-x+x)}{\cos x} d x \\ = & \int \frac{\sin [(n-1) x+x]}{\cos x} d x \\ & \int \frac{\sin (n-1) x \cos x+\cos (n-1) x \sin x}{\cos } d x\end{aligned}$
$$
\begin{aligned}
& =\int \sin (n-1) x d x+\int \frac{\cos (n-1) x \sin x}{\cos x} d x \\
& =\int \sin (n-1) x d x+\frac{1}{2} \int \frac{2 \sin x \cos (n-1) x}{\cos x} d x
\end{aligned}
$$
Since, $2 \sin x \cos y=+\sin (x+y)+\sin (x-y)$
$$
\begin{aligned}
& =-\frac{\cos (n-1) x}{(n-1)}+\frac{1}{2} \int \frac{\sin n x+\sin (2-n) x}{\cos x} d x \\
& =\frac{-\cos (n-1) x}{(n-1)} \\
& \quad+\frac{1}{2} \int \frac{\sin n x}{\cos x} d x-\frac{1}{2} \int \frac{\sin (n-2) x}{\cos x} d x
\end{aligned}
$$
From Eq. (i),
$$
\begin{aligned}
& \therefore I_n=-\frac{\cos (n-1) x}{(n-1)}+\frac{1}{2} I_n-\frac{1}{2} I_{n-2} \\
& \left(1-\frac{1}{2}\right) I_n=-\frac{\cos (n-1) x}{(n-1)}-\frac{1}{2} I_{n-2} \\
& I_n=\frac{-2}{n-1} \cos (n-1) x-I_{n-2}
\end{aligned}
$$