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

The integral from 0 to t of sin(x)f(t-x) is f(t-sin(t))

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

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

f (t - sin (t))

Solution steps

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

Take the constant out:

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

= f \cdot \int_{0}^{t} (t - x) sin (x) d x

Expand

(t - x) sin (x) : t sin (x) - x sin (x)

= f \cdot \int_{0}^{t} t sin (x) - x sin (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

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

\int_{0}^{t} t sin (x) d x = t (- cos (t) + 1)

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

= f (t (- cos (t) + 1) - (sin (t) - t cos (t)))

Expand

t (- cos (t) + 1) - (sin (t) - t cos (t)) : t - sin (t)

= f (t - sin (t))

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