Question

If the reduction formula for $I_n=\int \frac{\sin \mathrm{nx}}{\cos x}\mathrm{dx}$ is given by
$I_n+I_{n-2}=-\frac{2\cos \left\{\right(n-1\left)x\right\}}{n-1}$, then $\int \frac{\sin 3x}{\cos x}\mathrm{dx}$ is.

• A.
  $\cos 2x+\log \left(\sec x\right|+C$
• B.
  $\cos 2x-\log \left(\sec x\right|+C$
• C.
  $-\cos 2x-\log \left(\sec x\right|+C$
• D.
  $-\cos 2x+\log \left(\sec x\right|+C$

Answer 1

The correct option is C

$-\cos 2x-\log \left(\sec x\right|+C$
We know the reduction formula for $I_n=\int \frac{\sin \mathrm{nx}}{\cos x}\mathrm{dx}$ as :
$I_n+I_{n-2}=-\frac{2\cos \left\{\right(n-1\left)x\right\}}{n-1}$
Now, to find $I_n=\int \frac{\sin 3x}{\cos x}\mathrm{dx}$
we take $n=3$, thus substituting it back in the reduction formula we get
$I_3+I_1=-2\frac{\cos 2x}{2}$,
Here, $I_1=\int \frac{\mathrm{sinx}}{\mathrm{cosx}}\mathrm{dx}=\ln \left(\mathrm{secx}\right|+C$,
Thus we get $I_3=-\cos 2x-\ln \left(\mathrm{secx}\right|+C$
So, Option c. is correct.