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

The integral from 0 to t of sin(x)cos(t-x) is t/2 sin(t)

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integral from 0 to t of sin(x)cos(t-x)

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

\frac{t}{2} sin (t)

Solution steps

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

Use the following identity:

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

= \int_{0}^{t} \frac{sin ( x + t - x ) + sin ( x - ( t - 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}^{t} sin (x + t - x) + sin (x - (t - x)) d x

Apply u-substitution

: \int_{- t}^{t} \frac{sin ( t ) + sin ( u )}{2} d u

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

Take the constant out:

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

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

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

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

= \frac{1}{2} \cdot \frac{1}{2} (2 t sin (t) + 0)

Simplify

\frac{1}{2} \cdot \frac{1}{2} (2 t sin (t) + 0) : \frac{t}{2} sin (t)

= \frac{t}{2} sin (t)

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What is the integral from 0 to t of sin(x)cos(t-x) ?
The integral from 0 to t of sin(x)cos(t-x) is t/2 sin(t)