Question

If $n$ is a positive integer, and $I_n=\int \frac{\sin \left(nx\right)}{\cos x}dx$, then $I_n+I_{n-2}=$

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

Answer 1

Answer: (B)

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

$I_n=\int \frac{\sin \left(nx\right)}{\cos x}dx$
$I_{n-2}=\int \frac{\sin (n-2)}{\cos x}dx$
$I_n+I_{n-2}=\int \frac{\sin \left[nx\right]}{\cos x}dx+\int \frac{\sin \left[\right(n-2)x}{\cos x}dx$
Use $\sin (A+B)+\sin (A-B)=2\sin A\cos B$
$\therefore I_n+I_{n-2}=\int \frac{2\sin (n-1)x\cos x}{\cos x}dx$
$=2\int \sin (n-1)xdx=\frac{-2}{n-1}\cos (n-1)x+c$
$\Rightarrow I_n+I_{n-2}=\frac{-2}{(n-1)}\cos (n-1)x+c$