Question

Integrate: $\int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} $

Answer 1

Hint: First of all this is an indefinite integral which is one of the two types of integrals. Indefinite integrals have no limits on the integral, that is they are integrated indefinitely. Here in order to solve this particular given indefinite integral we use trigonometric sum to product formula and a few trigonometric identities and the trigonometric sum and difference formulas.

Step-By-Step answer:
The formulas used here are given below:
Trigonometric sum to product formula: $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$
Trigonometric cosine formula: $\cos 2A = 2{\cos ^2}A - 1$
Trigonometric cosine formula: $\cos 3A = 4{\cos ^3}A - 3\cos A$
$ \Rightarrow \int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} $
Now consider $\cos 5x + \cos 4x$, now apply sum to product formula here:
$ \Rightarrow \cos 5x + \cos 4x = 2\cos \left( {\dfrac{{5x + 4x}}{2}} \right)\cos \left( {\dfrac{{5x - 4x}}{2}} \right)$
$ \Rightarrow \cos 5x + \cos 4x = 2\cos \left( {\dfrac{{9x}}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)$
Now consider $\cos 3x$, apply the trigonometric identity:
$ \Rightarrow \cos 3x = 2{\cos ^2}\left( {\dfrac{{3x}}{2}} \right) - 1$
Substitute these solved expressions in the integral:
$ \Rightarrow \int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} = \int {\dfrac{{2\cos \left( {\dfrac{{9x}}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)}}{{1 - 2\left( {2{{\cos }^2}\left( {\dfrac{{3x}}{2}} \right) - 1} \right)}}} dx$
$ \Rightarrow \int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} = \int {\dfrac{{2\cos \left( {\dfrac{{9x}}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)}}{{3 - 4{{\cos }^2}\left( {\dfrac{{3x}}{2}} \right)}}} dx$
Now multiply and divide inside the integral with $\cos \dfrac{{3x}}{2}$, as given below:
$ \Rightarrow \int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} = \int {\dfrac{{2\cos \left( {\dfrac{{9x}}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)}}{{3 - 4{{\cos }^2}\left( {\dfrac{{3x}}{2}} \right)}}} \times \dfrac{{\cos \dfrac{{3x}}{2}}}{{\cos \dfrac{{3x}}{2}}}dx$
\[ \Rightarrow \int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} = \int {\dfrac{{2\cos \left( {\dfrac{{9x}}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)\cos \dfrac{{3x}}{2}}}{{3\cos \dfrac{{3x}}{2} - 4{{\cos }^3}\left( {\dfrac{{3x}}{2}} \right)}}} dx\]
As \[4{\cos ^3}\left( {\dfrac{{3x}}{2}} \right) - 3\cos \dfrac{{3x}}{2} = \cos \left( {3 \times \dfrac{{3x}}{2}} \right)\], is a trigonometric formula of cosine, simplifying the integral :
\[ \Rightarrow \int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} = \int {\dfrac{{2\cos \left( {\dfrac{{9x}}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)\cos \dfrac{{3x}}{2}}}{{ - \cos \left( {\dfrac{{9x}}{2}} \right)}}} dx\]
\[ \Rightarrow \int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} = - \int {2\cos \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{{3x}}{2}} \right)} dx\]
Now again applying the trigonometric product to sum formula here,
Consider \[2\cos \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{{3x}}{2}} \right) = \cos (2x) + \cos (x)\], substituting this in the integral:
\[ \Rightarrow \int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} = - \int {\left( {\cos (2x) + \cos (x)} \right)} dx\]
\[ \Rightarrow \int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} = - \int {\cos (2x)} dx - \int {\cos xdx} \]
\[ \Rightarrow \int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} = - \dfrac{{\sin (2x)}}{2} - \sin x + c\]


\[\therefore \int {\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}dx} = - \dfrac{{\sin (2x)}}{2} - \sin x + c\]

Note: Here totally three different kinds of trigonometric formulas are used to solve the given integral, which is an indefinite integral, which means that the integral has no limits, and is integrated indefinitely. Here the trigonometric sum to product formula is used and also the trigonometric product to sum formula is used. Various trigonometric identities are also used which are very important to solve any trigonometric based problem.