Question

$\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}dx$

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

Answer 1

The correct option is C $-(\sin x+\frac{1}{2}\sin 2x)$

Let $I=\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}dx$
Multipling numerator and denominator by $\sin 3x$
$I=\int \frac{\left(2\sin \frac{3x}{2}\cos \frac{3x}{2}\right)\left(2\cos \frac{9x}{2}\cos \frac{x}{2}\right)}{\sin 3x-\sin 6x}dx$
$D^{\prime }=\sin 3x-\sin 6x=-2\sin \frac{3x}{2}\cos \frac{9x}{2}$
$I=\int -\left(2\cos \frac{3x}{2}\cos \frac{x}{2}\right)dx=-\int (\cos x+\cos 2x)dx$
$=-(\sin x+\frac{1}{2}\sin 2x)$
Hence, option 'C' is correct.