What is integral from 0 to pi of f(x)sin(x) ?

The answer to integral from 0 to pi of f(x)sin(x) is fpi

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integral from 0 to pi of f(x)sin(x)

\int_{0}^{π} f (x) sin (x) d x

f π

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

= f [- x cos (x) + sin (x)]_{0}^{π}

Compute the boundaries:

π

= f π

\int \frac{3}{5 - 4 u - u ^{2}} d u

9 y^{^{''}} + 6 y^{^{'}} + 4 y = 0

\int 2 e^{t} d t

7 cos (9 x), 7 - 7 cos (9 x), 0, \frac{π}{9}

x y^{^{'}} + 2 y = x^{2}

What is integral from 0 to pi of f(x)sin(x) ?
The answer to integral from 0 to pi of f(x)sin(x) is fpi