What is the integral of (cos(ax)-cos(bx))/(x^2) ?

The integral of (cos(ax)-cos(bx))/(x^2) is -(cos(ax)-cos(bx))/x-a\Si(ax)+b\Si(bx)+C

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integral of (cos(ax)-cos(bx))/(x^2)

\int \frac{cos ( a x ) - cos ( b x )}{x ^{2}} d x

- \frac{cos ( a x ) - cos ( b x )}{x} - a Si (a x) + b Si (b x) + C

Solution steps

\int \frac{cos ( a x ) - cos ( b x )}{x ^{2}} d x

Apply Integration By Parts

: - \frac{cos ( a x ) - cos ( b x )}{x} - \int - \frac{- a sin ( a x ) + b sin ( b x )}{x} d x

= - \frac{cos ( a x ) - cos ( b x )}{x} - \int - \frac{- a sin ( a x ) + b sin ( b x )}{x} d x

\int - \frac{- a sin ( a x ) + b sin ( b x )}{x} d x = a Si (a x) - b Si (b x)

= - \frac{cos ( a x ) - cos ( b x )}{x} - (a Si (a x) - b Si (b x))

Simplify

= - \frac{cos ( a x ) - cos ( b x )}{x} - a Si (a x) + b Si (b x)

Add a constant to the solution

= - \frac{cos ( a x ) - cos ( b x )}{x} - a Si (a x) + b Si (b x) + C

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What is the integral of (cos(ax)-cos(bx))/(x^2) ?
The integral of (cos(ax)-cos(bx))/(x^2) is -(cos(ax)-cos(bx))/x-a\Si(ax)+b\Si(bx)+C