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

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

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

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

f (x - sin (x))

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: [t cos (x - t) - \int cos (x - t) d t]_{0}^{x}

= f [t cos (x - t) - \int cos (x - t) d t]_{0}^{x}

\int cos (x - t) d t = - sin (x - t)

= f [t cos (x - t) - (- sin (x - t))]_{0}^{x}

Simplify

= f [t cos (x - t) + sin (x - t)]_{0}^{x}

Compute the boundaries:

x - sin (x)

= f (x - sin (x))

\int_{- 1}^{8} 1 + x d x

\int_{0}^{3} x d x

\int_{1}^{9} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

\int_{0}^{2} ∣ 5 t cos (π t) ∣ d t

\int_{1}^{2} (x - 3) (x + 6) d x

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