Question

The value of $\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}dx$, is equal to

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

Answer 1

Answer: (D)

$\sin x+\frac{1}{2}\sin 2x+c$

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

Apply linearity :

$-\int \frac{\cos \left(5x\right)+\cos \left(4x\right)dx}{2\cos \left(3x\right)-1}......\left(A\right)$

simplify using trigonometric identifies

$=\int \left(\cos 2x\right)+\left(\cos x\right)dx$

$=\int \cos \left(2x\right)dx+\int \cos \left(x\right)dx$

$\int \cos \left(2x\right)dx\Rightarrow$ substitute $u=2x$

$dx=\frac{1}{2}du=\frac{1}{2}\int \cos \left(u\right)du$

$\int \cos \left(u\right)du\Rightarrow \sin \left(u\right)$

$\therefore \frac{1}{2}\int \cos \left(u\right)du=\frac{\sin \left(u\right)}{2}=\frac{\sin \left(2x\right)}{2}$

$\int \cos xdx=\sin \left(x\right)$

$\therefore \int \cos \left(2x\right)dx+\int \cos \left(x\right)dx$

$=\frac{\sin \left(2x\right)}{2}+\sin \left(x\right)$

$=\frac{-\sin \left(2x\right)}{2}+\sin \left(x\right)$

$=-\frac{\sin \left(2x\right)}{2}-\sin \left(x\right)+c$....... from $\left(A\right)$

$=\sin \left(x\right)+\frac{1}{2}\sin \left(2x\right)+c$