1.

Solve`int` `(cos 4x-cos 2x)/(sin 4x- sin 2x)` dx

Answer» Let `I = int (cos4-cos2x)/(sin4x-sin2x)dx`
`= int (-2sin3xsinx)/(2cos3xsinx) dx`
`= int (sin3x)/(cos3x) dx`
Let, `cos3x = t => -3sin3xdx = dt`
So, our integral becomes,
`I = -1/3 int dt/t`
`=>I = -1/3 log|cos3x|+c`


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