How to Evaluate this Integral $\int \frac{\cos 6x+\cos 8x}{1+2\cos 5x}dx$?

Question

Evaluate the following integral

$$I=\int \frac{\cos 6x+\cos 8x}{1+2\cos 5x}dx$$

My work

I tried to change $\cos x$ into $\sin x$ and try to substitute. But I am not getting it ...

Answer 1

Here, I will try to explain the process of integration. I did the computations in MAPLE so even don't think of doing this lengthy calculations by hand! :) You can download the MAPLE file from here.

$1.$ Convert multiple angels to single angles using the following identities

$$\eqalign{ & \cos \left( {8x} \right) = 128\cos {\left( x \right)^8} - 256\cos {\left( x \right)^6} + 160\cos {\left( x \right)^4} - 32\cos {\left( x \right)^2} + 1 \cr & \cos \left( {6x} \right) = 32\cos {(x)^6} - 48\cos {(x)^4} + 18\cos {(x)^2} - 1 \cr & \cos \left( {5x} \right) = 16\cos {(x)^5} - 20\cos {(x)^3} + 5\cos (x) \cr} $$

and rewrite the integrand

$$f = {{128\cos {{\left( x \right)}^8} - 224\cos {{\left( x \right)}^6} + 112\cos {{\left( x \right)}^4} - 14\cos {{\left( x \right)}^2}} \over {1 + 32\cos {{\left( x \right)}^5} - 40\cos {{\left( x \right)}^3} + 10\cos \left( x \right)}}$$

$2.$ Use the tangent half angle substitution

$$\eqalign{ & u = \tan \left( {{x \over 2}} \right) \cr & dx = {2 \over {1 + {u^2}}}du \cr & \cos \left( x \right) = {{1 - {u^2}} \over {1 + {u^2}}} \cr} $$

Hence your integral will be

$$\eqalign{ & g(u) = f\left( {x(u)} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {{ - 4{u^{16}} + 368{u^{14}} - 4368{u^{12}} + 16016{u^{10}} - 24024{u^8} + 16016{u^6} - 4368{u^4} + 368{u^2} - 4} \over {\left( {{u^2} - 3} \right){{\left( {{u^2} + 1} \right)}^4}\left( {{u^8} - 92{u^6} + 134{u^4} - 28{u^2} + 1} \right)}} \cr & h(u) = g(u){2 \over {1 + {u^2}}} \cr & I = \int {f(x)dx} = \int {h(u)du} \cr} $$

$3.$ The hard part is to use partial fractions to decompose $h(u)$ into

$$\eqalign{ & h(u) = \mathop \sum \limits_{R = RootOf\left( {{x^2} - 3} \right)} {R \over {30\left( {u - R} \right)}} + \mathop \sum \limits_{R = RootOf\left( {{x^2} + 1} \right)} - {2 \over {{{\left( {u - R} \right)}^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\, + \mathop \sum \limits_{R = RootOf\left( {{x^2} + 1} \right)} - {{4R} \over {{{\left( {u - R} \right)}^3}}} + \mathop \sum \limits_{R = RootOf\left( {{x^2} + 1} \right)} {4 \over {{{\left( {u - R} \right)}^4}}} \cr & \,\,\,\,\,\,\,\,\,\,\, + \mathop \sum \limits_{R = RootOf\left( {{x^8} - 92{x^6} + 134{x^4} - 28{x^2} + 1} \right)} {{ - {{71} \over {1920}}{R^7} + {{1307} \over {384}}{R^5} - {{3263} \over {640}}{R^3} + {{2293} \over {1920}}R} \over {u - R}} \cr} $$

$4.$ Now, integrate the decomposed $h(u)$. As you can see, the last term with eighth degree polynomial which does have 8 real distinct roots that are hard to compute is a disaster. This makes the final result a little ugly given by MAPLE or Wolfram-Alpha! So sometimes to compute an integral you may give those computing monsters some directions manually to obtain a more beautiful and tidier final result!

$$\eqalign{ & I = \mathop \sum \limits_{R = RootOf\left( {{x^2} - 3} \right)} {{R\ln \left( {u - R} \right)} \over {30}} + \mathop \sum \limits_{R = RootOf\left( {{x^2} + 1} \right)} {2 \over {u - R}} \cr & \,\,\, + \mathop \sum \limits_{R = RootOf\left( {{x^2} + 1} \right)} {{2R} \over {{{\left( {u - R} \right)}^2}}} + \mathop \sum \limits_{R = RootOf\left( {{x^2} + 1} \right)} - {4 \over {3{{\left( {u - R} \right)}^3}}} \cr & \,\,\, + \mathop \sum \limits_{R = RootOf\left( {{x^8} - 92{x^6} + 134{x^4} - 28{x^2} + 1} \right)} \left( { - {{71} \over {1920}}{R^7} + {{1307} \over {384}}{R^5} - {{3263} \over {640}}{R^3} + {{2293} \over {1920}}R} \right)\ln \left( {u - R} \right) \cr} $$

As you could see, the integrals to be evaluated after partial fraction decomposition were so easy; however, the process was frustrating and brutal! :)