Practicing Success
$\int \frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x} d x$ is equal to : |
$\frac{\sin 2 x}{2}-\sin x+c$ $-\frac{\sin 2 x}{2}+\sin x+c$ $-\frac{\sin 2 x}{2}-\sin x+c$ $\frac{\sin 2 x}{2}+\sin x+c$ |
$-\frac{\sin 2 x}{2}-\sin x+c$ |
$\int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2} \cos \frac{3 x}{2}}{\left[1-2\left(2 \cos 2 \frac{3 x}{2}-1\right)\right] \cos \frac{3 x}{2}} d x=\int \frac{2 \cos \frac{9 x}{2} \cos \frac{3 x}{2} \cos \frac{x}{2}}{3 \cos \frac{3 x}{2}-4 \cos ^3 \frac{3 x}{2}} d x$ $=\int \frac{2 \cos \frac{9 x}{2} \cos \frac{3 x}{2} \cos \frac{x}{2}}{-\cos \frac{9 x}{2}} d x=-\int(\cos 2 x+\cos x) d x$ $=-\frac{\sin 2 x}{2}-\sin x+c$ Hence (3) is the correct answer. |