If value of integral is given, then find $k$

Question

If

$$\int \frac{\cos (6x) +\cos(9x)}{1-2 \cos(5x)}dx=- \frac{\sin(4x)}{k}-\sin(x)+C$$

then find value of $k$

Could someone give me hint as how to initiate the solution of given integral?

Answer 1

$$\int \frac{\cos (6x) +\cos(9x)}{1-2 \cos(5x)}dx=- \frac{\sin(4x)}{k}-\sin(x)+C$$

If $k$ is a constant, differentiating the equation gives us:

$$ \frac{\cos (6x) +\cos(9x)}{1-2 \cos(5x)}=- \frac{4\cos(4x)}{k}-\cos(x)$$

Now set $x=0$ and solve for the value of $k$.

Answer 2

$$\cos2A+\cos(6x-2A)=2\cos3x\cos(3x-2A)$$

$$1-2\cos2x=1-2(2\cos^2x-1)=-\frac{\cos3x}{\cos x}$$

Finally $2\cos C\cos D=\cos(C+D)+\cos(C-D)$

Answer 3

let $\displaystyle \int \frac{\cos 6x - \cos 9 x}{1+2\cos 5x}dx = \int \frac{(\cos 6x+\cos 4x)-(\cos 9x+\cos x)-\cos 4x+\cos x}{1+2\cos 5x}dx$

$\displaystyle I = \int\frac{2\cos 5x \cos x+\cos x-2\cos5x\cos 4x-\cos 4x }{1+2\cos 5x}dx$

$\displaystyle = \int\frac{\cos x(2\cos 5x +1)-\cos 4x(2\cos 5x+1)}{1+2\cos 5x}dx = \int (\cos x-\cos 4x)dx$

$\displaystyle = \sin x-\frac{\sin 4x}{4}+\mathcal{C}$