![Find the integral of sinxcosxcos2x dx
Solution
Find `int (sinx cosx cos2x) dx`
(1) Let `u=cos2x` . Then `(du)=-2sin(2x)=-4sinxcosxdx`
(2) Substituting we get:
`int(sinxcosxcos(2x))dx=-1/4 int u du`
`=-1/4[u^2/2+C_1]`
`=-1/4[(cos^2(2x))/2+C_1]`
`=-1/8(cos^2(2x))+C`
`=-1/16 cos(4x)+C`
Thus the solution is `-1/16 cos(4x)+C`
** `d/(dx)[-1/16 cos(4x)+C]=-1/16(4)(-sin4x)`
`=1/4sin4x`
`=1/4[2sin2xcos2x]`
`=1/4[4sinxcosxcos2x]`
`=sinxcosxcos2x` as required. **](https://image.slidesharecdn.com/findtheintegralofsinxcosxcos2xdxsolutionfindintsinxc-230402072042-b1d322b0/85/Find-the-integral-of-sinxcosxcos2x-dxSolutionFind-int-sinx-c-pdf-1-320.jpg)
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Find the integral of sinxcosxcos2x dxSolutionFind `int (sinx c.pdf
- 1. Find the integral of sinxcosxcos2x dx Solution Find `int (sinx cosx cos2x) dx` (1) Let `u=cos2x` . Then `(du)=-2sin(2x)=-4sinxcosxdx` (2) Substituting we get: `int(sinxcosxcos(2x))dx=-1/4 int u du` `=-1/4[u^2/2+C_1]` `=-1/4[(cos^2(2x))/2+C_1]` `=-1/8(cos^2(2x))+C` `=-1/16 cos(4x)+C` Thus the solution is `-1/16 cos(4x)+C` ** `d/(dx)[-1/16 cos(4x)+C]=-1/16(4)(-sin4x)` `=1/4sin4x` `=1/4[2sin2xcos2x]` `=1/4[4sinxcosxcos2x]` `=sinxcosxcos2x` as required. **