If I_n = int frac{sin nx}{cos x} dx,then I_n | vedclass

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(A) Given $I_n = \int \frac{\sin nx}{\cos x} dx$  ... $(i)$Consider $I_n - I_{n-2} = \int \frac{\sin nx - \sin (n-2)x}{\cos x} dx$Using the formula $\

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$\frac{-2}{n-1} \cos (n-1)x - I_{n-2}$

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(A) Given $I_n = \int \frac{\sin nx}{\cos x} dx$  ... $(i)$Consider $I_n - I_{n-2} = \int \frac{\sin nx - \sin (n-2)x}{\cos x} dx$Using the formula $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$:$I_n - I_{n-2} = \int \frac{2 \cos \left(\frac{nx + nx - 2x}{2}\right) \sin \left(\frac{nx - nx + 2x}{2}\right)}{\cos x} dx$$I_n - I_{n-2} = \int \frac{2 \cos (n-1)x \sin x}{\cos x} dx$$I_n - I_{n-2} = 2 \int \cos (n-1)x dx$$I_n - I_{n-2} = 2 \left[ \frac{\sin (n-1)x}{n-1} \right] + C$Alternatively,using $I_n + I_{n-2} = \int \frac{\sin nx + \sin (n-2)x}{\cos x} dx = \int \frac{2 \sin (n-1)x \cos x}{\cos x} dx = 2 \int \sin (n-1)x dx = -\frac{2 \cos (n-1)x}{n-1} + C$Thus,$I_n = \frac{-2}{n-1} \cos (n-1)x - I_{n-2}$.

If $I_n = \int \frac{\sin nx}{\cos x} dx$,then $I_n =$

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