Intuition behind the multiplication and possibility of an alternative method

Question

Evaluate, with a practical approach, $$\int\frac{\cos(9x)+\cos(6x)}{2\cos(5x)-1}dx$$

The solution that I have first multiplies both numerator and denominator by $\cos\left(\frac{5x}{2}\right)$ then uses the famous formula for $\cos C+\cos D$. Then the integral is reduced to a very basic one.

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My question is how did we just randomly multiplied both numerator and denominator by that $\cos\left(\frac{5x}{2}\right)$. Is there some reason behind it? Or is it cause the question is framed in such a way?

Also, is there any other alternative (elementary) method to this?

Any help is greatly appreciated.

Answer 1

Using $$ \cos(A)+\cos(B)=2\cos\bigg(\frac{A+B}2\bigg)\cos\bigg(\frac{A-B}2\bigg), \cos(2\theta)=2\cos^2(\theta)-1 $$ and $$ \cos(3\theta)=4\cos^3(\theta)-3\cos(\theta), $$ one has \begin{eqnarray} &&\int\frac{\cos(9x)+\cos(6x)}{2\cos(5x)-1}dx\\ &=&\int\frac{2\cos(\frac{15}2x)\cos(\frac32x)}{4\cos^2(\frac52x)-3}dx\\ &=&\int\frac{2(4\cos^3(\frac52x)-3\cos(\frac52x))\cos(\frac32x)}{4\cos^2(\frac52x)-3}dx\\ &=&\int2\cos(\frac52x)\cos(\frac32x)dx\\ &=&\int(\cos(4x)+\cos(x))dx\\ &=&\frac14\sin(4x)+\sin(x)+C. \end{eqnarray}

Answer 2

Note that \begin{align} &\cos9x +\cos x= 2\cos 5x \cos 4x \\ &\cos6x+\cos 4x =2\cos 5x \cos x \end{align} which lead to $$ \cos9x +\cos 6x= (2\cos 5x -1)(\cos x+\cos 4x)$$ and \begin{align}\int\frac{\cos 9x+\cos6x}{2\cos5x-1}dx =\int(\cos x+\cos 4x)dx=\sin x+\frac14\sin 4x \end{align}

Answer 3

This question is really more about trigonometry than calculus.

Denoting $z := e^{ix}$, so that $$\cos k x = \frac{1}{2}(z^k + z^{-k}) ,$$ the integrand is \begin{multline}\frac{\frac12 (z^9 + z^{-9}) + \frac12 (z^6 + z^{-6})}{2 \cdot \frac12 (z^5 - z^{-5}) - 1} = \frac{z^{18} + z^{15} + z^3 + 1}{2 z^4 (z^{10} - z^5 + 1)} \\ = \frac{z^{18} + z^{15} + z^3 + 1}{2 z^4 (z^{10} - z^5 + 1)} = \frac{(z^{15} + 1) (z^3 + 1)}{2 z^4 (z^{10} - z^5 + 1)} = \frac{(z^{10} - z^5 + 1) (z^5 + 1) (z^3 + 1)}{2 z^4 (z^{10} - z^5 + 1)}, \end{multline} where the last equality uses the factorization of $z^{15} + 1$ as a sum of perfect cubes. Canceling leaves $$\frac{1}{2}(z^4 + z + z^{-1} + z^{-4}) = \frac{1}{2} (z^4 + z^{-4}) + \frac{1}{2} (z + z^{-1}) = \boxed{\cos 4 x + \cos x} .$$

Answer 4

One can multiply num/den by $\sin(5x)$, because one can note the double angle formula of sine and the difference of sines in the denominator: $$\int\frac{\cos(9x)+\cos(6x)}{2\cos(5x)-1}dx=\int\frac{[\cos(9x)+\cos(6x)]\cdot \sin(5x)}{\sin(10x)-\sin(5x)}dx=\\ \int\frac{2\cos(\frac32x)\cos(\frac{15}2x)\sin(5x)}{2\cos(\frac{15}2x)\sin(\frac52x)}dx=\int2\cos(\frac32x)\cos(\frac52x)dx=\\ \int \cos4x+\cos x \ dx=\frac14 \sin 4x+\sin x+C $$ Note: The following formulas are used: $$ \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}\\ \sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}\\ \sin(2A)=2\sin A\cos A$$