What I've done is this:$$\int\dfrac{\cos5x+\cos4x}{1-2\cos3x}{dx}$$ $$\int \dfrac{\sin 3x}{\sin 3x}\left[\dfrac{\cos5x+\cos4x}{1-2\cos3x}\right]{dx}$$ $$\dfrac {1}{2}\int\dfrac{\sin 8x -\sin 2x +\sin 7x -\sin x}{\sin 3x - \sin 6x}$$ $$-\dfrac {1}{2}\int\dfrac{ \sin \frac{7x}{2} +\sin \frac{5x}{2} } {\sin \frac{3x}{2} }$$ $$-\int\dfrac{ \sin {3x}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$ $$-\int\dfrac{2\sin \frac{3x}{2} \cos \frac{3x}{2}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$ $$-\int {2\cos \frac{3x}{2}\cos \frac{x}{2} }$$ $$ -\left(\frac{\sin 2x}{2} +\sin x \right) +c $$ Is there any other way to do so ? Is it possible to do it by substitution ?


Is there any other way to evaluate it ?

Hint. One may observe that $$ \frac{\cos(5x)+\cos(4x)}{1-2\cos(3x)}=-\cos(2x)-\cos (x) $$ then the evaluation is easier.

Edit. Here is a way to obtain such a simplification. One may set $u=e^{ix}$ then using De moivre's formula for $\cos(\cdot)$ one gets $$ \begin{align} \frac{\cos(5x)+\cos(4x)}{1-2\cos(3x)}&=\frac{\dfrac{u^5+\dfrac1{u^5}}2+\dfrac{u^4+\dfrac1{u^4}}2}{1-\big(u^3+\dfrac1{u^3}\big)} \\\\&=-\frac12\:\frac{1+u+u^3+u^4}{u^2} \\\\&=- \dfrac{u^2+\dfrac1{u^2}}2-\dfrac{u+\dfrac1{u}}2 \\\\&=-\cos(2x)-\cos (x). \end{align} $$

  • $\begingroup$ Clever and instructive approach to use De Moivre backstage! (+1) $\endgroup$ – Markus Scheuer Jul 21 '16 at 19:07
  • $\begingroup$ @AakashKumar By first multiplying numerator and denominator by $u^5$ one gets $\frac{u^{10}+u^9+u+1}{u^5-u^8+u^2}=\frac{u(u^9+1)+(u^9+1)}{\frac{u^9+1}{u^3+1} \cdot u^2}$ giving $\frac{(u+1)(u^3+1)}{u^2}$ thus $\frac{1+u+u^3+u^4}{u^2}$. Tell me if it is Ok for you. $\endgroup$ – Olivier Oloa Jul 22 '16 at 8:24

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