What is the integral from 0 to 1 of cos(ax)cos(bx) ?

The integral from 0 to 1 of cos(ax)cos(bx) is 1/2 ((sin(a+b))/(a+b)+(sin(a-b))/(a-b))

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integral from 0 to 1 of cos(ax)cos(bx)

\int_{0}^{1} cos (a x) cos (b x) d x

\frac{1}{2} (\frac{sin ( a + b )}{a + b} + \frac{sin ( a - b )}{a - b})

Solution steps

\int_{0}^{1} cos (a x) cos (b x) d x

Rewrite using trig identities

= \int_{0}^{1} \frac{cos ( a x + b x ) + cos ( a x - b x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{1} cos (a x + b x) + cos (a x - b x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{1} cos (a x + b x) d x + \int_{0}^{1} cos (a x - b x) d x)

\int_{0}^{1} cos (a x + b x) d x = \frac{sin ( a + b )}{a + b}

\int_{0}^{1} cos (a x - b x) d x = \frac{sin ( a - b )}{a - b}

= \frac{1}{2} (\frac{sin ( a + b )}{a + b} + \frac{sin ( a - b )}{a - b})

\int_{- 2}^{1} \frac{x}{x ^{2} + 1} d x

\int_{0}^{2} 10 + 2 x - x^{2} d x

\int_{p}^{a + p} x d x

\int_{- 1}^{0} x + x^{2} d x

\int_{0}^{6} 5 x d x

What is the integral from 0 to 1 of cos(ax)cos(bx) ?
The integral from 0 to 1 of cos(ax)cos(bx) is 1/2 ((sin(a+b))/(a+b)+(sin(a-b))/(a-b))