Question

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

• A. $-\frac{2}{(n-1)}\cos (n-1)x+I_{n-2}$
• B. $\frac{2}{(n-1)}\cos (n-1)x+I_{n-2}$
• C. $\frac{2}{(n-1)}\sin (n-1)x-I_{n-2}$
• D. $-\frac{2}{(n-1)}\sin (n-1)x+I_{n-2}$

Answer 1

Answer: (C)

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

$I_n=\int \frac{cosnx}{cosx}dx,In-2=\int \frac{cos(n-2)x}{cosx}dx$
$I_n+I_n-2=\int \frac{cosdx}{cosx}dx+\int \frac{cos(n-2)x}{cosx}dx$
$\int \frac{cos\left[\right(n-1)x+x]}{cosx}dx+\int \frac{cos\left[\right(n-1)x-x]}{cosx}dx$
$\int \frac{2cos(n-1)xcosx}{cosx}dx=2\int cos(n-1)xdx$
$\frac{2}{n-1}sin(n-1)x+c$
$I_n+I_{n-2}=\frac{2}{n-1}x+c$
$I_n=\frac{2}{n-1}sin(n-1)x-I_{n-2}$