If $I_n=\int \frac{\sin n x}{\cos x} d x$, then $I_n=$

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Aicharya · Accepted Answer

$\begin{aligned}= & \int \frac{\sin (n x-x+x)}{\cos x} d x \\ = & \int \frac{\sin [(n-1) x+x]}{\cos x} d x \\ & \int \frac{\sin (n-1) x \cos x+\cos (n-1) x \sin x}{\cos } d x\end{aligned}$ $ \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} $

If $I_n=\int \frac{\sin n x}{\cos x} d x$, then $I_n=$

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