What is the integral of (x-1)sin(2-x) ?

The integral of (x-1)sin(2-x) is xcos(-x+2)-cos(-x+2)+sin(-x+2)+C

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integral of (x-1)sin(2-x)

\int (x - 1) sin (2 - x) d x

x cos (- x + 2) - cos (- x + 2) + sin (- x + 2) + C

Solution steps

\int (x - 1) sin (2 - x) d x

Apply u-substitution

: \int u sin (- u + 1) d u

= \int u sin (- u + 1) d u

Apply u-substitution

: \int - (- v + 1) sin (v) d v

= \int - (- v + 1) sin (v) d v

Take the constant out:

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

= - \int (- v + 1) sin (v) d v

Apply Integration By Parts

: - cos (v) (- v + 1) - \int cos (v) d v

= - (- cos (v) (- v + 1) - \int cos (v) d v)

\int cos (v) d v = sin (v)

= - (- cos (v) (- v + 1) - sin (v))

Substitute back

= - (- cos (- (x - 1) + 1) (- (- (x - 1) + 1) + 1) - sin (- (x - 1) + 1))

Simplify

- (- cos (- (x - 1) + 1) (- (- (x - 1) + 1) + 1) - sin (- (x - 1) + 1)) : x cos (- x + 2) - cos (- x + 2) + sin (- x + 2)

= x cos (- x + 2) - cos (- x + 2) + sin (- x + 2)

Add a constant to the solution

= x cos (- x + 2) - cos (- x + 2) + sin (- x + 2) + C

\int \frac{1}{x ^{2} + 16 x + 73} d x

\int \frac{x ^{2} - 4 x + 4}{x ^{2} - 4 x + 13} d x

\int (x^{2})^{'} d x

\int x^{3} ln (x^{2} + 3) d x

\int - cos (- x) d x

What is the integral of (x-1)sin(2-x) ?
The integral of (x-1)sin(2-x) is xcos(-x+2)-cos(-x+2)+sin(-x+2)+C