Find $$\int \cos{x} \cos{2x} \cos{3x}\;dx $$
This question is from Basic Mathematics. Please explain how I can solve it according to class 11th student.
My work is below: $ \implies \displaystyle\int\cos{x} \cos{2x} \cos{3x}\, dx\\ \implies\displaystyle\int\cos{x}(2\cos^2x - 1)(4\cos^3x - 3\cos{x}) \, dx\\ \implies\displaystyle\int\cos{x}(8\cos^5x-6\cos^3x-4\cos^3x+3\cos{x}) \, dx\\ \implies\displaystyle\int(8\cos^6x-6\cos^4x-4\cos^4x+3\cos^2x) \, dx\\ \implies\displaystyle\int(8\cos^6x-10\cos^4x+3\cos^2x) \, dx\\ \implies8\displaystyle\int\cos^6x \, dx-10\int\cos^4x \, dx+3\int\cos^2x \, dx\\ $
Now I'm unable to figure out what should I do next. I mean I know that $\displaystyle\int\cos{x}\,dx = \sin{x} + C$ but because of exponents I'm stuck here.
It is much simpler to use the formula $\cos \, A \cos \, B =\frac {\cos (A+B) +\cos (A-B)} 2$.
You should get $\frac 1 4 (1+\cos (6x)+\cos (4x)+\cos(2x))$ which is easy to integrate.