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?
If value of integral is given, then find $k$
2
$\begingroup$
calculus
integration
indefinite-integrals
3 Answers
6
$$\int \frac{\cos (6x) +\cos(9x)}{1-2 \cos(5x)}dx=- \frac{\sin(4x)}{k}-\sin(x)+C$$
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 $k$.
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0Is there any simple way to integrate $L.H.S.$? – 2017-01-21
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0@Mathgeek That sounds like a great *new* question to ask on stack exchange. :) – 2017-01-21
1
$$\cos2A+\cos(6x-2A)=2\cos3x\cos(3x-2A)$$
$$1-2\cos2x=1-2(2\cos^2x-1)=-\frac{\cos3x}{\cos x}$$
Finally $2\cos C\cos D=\cos(C+D)+\cos(C-D)$
0
let $\displaystyle \int \frac{\cos 6x - \cos 9 x}{1+2\cos 5x}dx = \int \frac{(\cos 6x+\cos 4x)-(\cos 9x+\cos x)-\cos 4x+\cos x}{1+2\cos 5x}dx$
$\displaystyle I = \int\frac{2\cos 5x \cos x+\cos x-2\cos5x\cos 4x-\cos 4x }{1+2\cos 5x}dx$
$\displaystyle = \int\frac{\cos x(2\cos 5x +1)-\cos 4x(2\cos 5x+1)}{1+2\cos 5x}dx = \int (\cos x-\cos 4x)dx$
$\displaystyle = \sin x-\frac{\sin 4x}{4}+\mathcal{C}$