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Find the integrals of the functions\[\cos2x\;\cos4x\;\cos6x\]

1 Answer

  • $(i)2\cos x\cos y=\cos (x+y)+\cos (x-y).$
  • $(ii)\frac{1+\cos 2x}{2}=\cos^2x.$
Given $I=\int\cos 2x\cos 4x\cos 6xdx.$
Using the information in the tool box we can write,
$I=\frac{1}{2}\int (\cos 6x+\cos 2x)\cos 6x.dx.$
On multiplying we get,
$\;\;=\frac{1}{2}\int \cos^26xdx+\frac{1}{2}\int \cos 2x\cos 6x dx.$
$\;\;\;=\frac{1}{2}\int \cos 2x\cos 6x dx=\frac{1}{2}\int[\cos (2x+6x)+\cos (2x-6x)]dx.$
$\;\;\;=\frac{1}{2}\int\cos ^26xdx+\frac{1}{2}\int\frac{1}{2}(\cos 8x+\cos 4x)dx.$
But $\cos^2x=\frac{(1+\cos 2x}{2},$so $\cos^2{6x}=\frac{(1+\cos 12x)}{2}$.
$\;\;\;=\frac{1}{2}\int\frac{1+\cos 12x}{2}dx+\frac{1}{4}\int\cos 8xdx+\frac{1}{4}\int\cos 4xdx.$
On separating the terms we get,
$\;\;\;=\frac{1}{4}\int dx+\frac{1}{4}\int\cos 12x dx+\frac{1}{4}\int \cos 8x dx+\frac{1}{4}\cos 4x dx.$
On integrating we get,
$\;\;\;=\frac{1}{4}x+\frac{1}{4}.\frac{1}{12}\sin12x+\frac{1}{4}\frac{1}{8}\sin 8x+\frac{1}{4}\frac{1}{4}\sin 4x+c.$
Taking $\frac{1}{4}$ as the common factor,
$\;\;\;=\frac{1}{4}[x+\frac{1}{12}\sin 12x+\frac{1}{8}\sin 8x+\frac{1}{4}\sin 4x]+c.$


answered Jan 30, 2013 by sreemathi.v