What is the integral from 0 to t of cos(u)cos(t-u) ?

The integral from 0 to t of cos(u)cos(t-u) is 1/2 (tcos(t)+sin(t))

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integral from 0 to t of cos(u)cos(t-u)

\int_{0}^{t} cos (u) cos (t - u) d u

\frac{1}{2} (t cos (t) + sin (t))

Solution steps

\int_{0}^{t} cos (u) cos (t - u) d u

Factor

t - u : - (- t + u)

= \int_{0}^{t} cos (- (- t + u)) cos (u) d u

Simplify

cos (- (- t + u)) cos (u) : cos (- t + u) cos (u)

= \int_{0}^{t} cos (- t + u) cos (u) d u

Rewrite using trig identities

= \int_{0}^{t} \frac{1}{2} (cos (- t + 2 u) + cos (- t)) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{t} cos (- t + 2 u) + cos (- t) d u

Simplify

cos (- t + 2 u) + cos (- t) : cos (t) + cos (- t + 2 u)

= \frac{1}{2} \cdot \int_{0}^{t} cos (t) + cos (- t + 2 u) d u

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}^{t} cos (t) d u + \int_{0}^{t} cos (- t + 2 u) d u)

\int_{0}^{t} cos (t) d u = t cos (t)

\int_{0}^{t} cos (- t + 2 u) d u = sin (t)

= \frac{1}{2} (t cos (t) + sin (t))

What is the integral from 0 to t of cos(u)cos(t-u) ?
The integral from 0 to t of cos(u)cos(t-u) is 1/2 (tcos(t)+sin(t))