Question

The anti-derivative of $\frac{\cos 5x+\cos 4x}{1-2\cos 3x}$ is

• A. $\frac{\sin 2x}{2}+\cos x+c$
• B. $-\frac{\sin 2x}{2}+\sin x+c$
• C. $-\frac{\sin 2x}{2}-\sin x+c$
• D. $\frac{\sin 2x}{2}-\cos x+c$

Answer 1

The correct option is C $-\frac{\sin 2x}{2}-\sin x+c$

The given anti-derivative $=\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}dx$

$=\int \frac{2\cos \frac{9x}{2}\cos \frac{x}{2}}{1-2\left\{2{\cos }^2\frac{3x}{2}-1\right\}}dx=\int \frac{2\cos \frac{9x}{2}\cos \frac{3x}{2}}{3-4{\cos }^2\frac{3x}{2}}$

$=\int \frac{2\cos \frac{9x}{2}\cos \frac{x}{2}\cos \frac{3x}{2}}{3\cos \frac{3x}{2}-4{\cos }^3\frac{3x}{2}}dx$

$[$Multiplying and dividing by $\frac{3x}{2}]$

$=\int \frac{2\cos \frac{9x}{2}\cos \frac{3x}{2}\cos \frac{x}{2}}{-\cos \frac{9x}{2}}dx$

$[\cos 3x=4{\cos }^{3}x-3\cos x]$

$=-\int 2\cos \frac{3x}{2}\cos \frac{x}{2}dx=-\int \left(\cos 2x+\cos x\right)dx$

$=-\frac{\sin 2x}{2}-\sin x+c$