What is the integral from 0 to 5 of sin(-1+x) ?

The integral from 0 to 5 of sin(-1+x) is -cos(4)+cos(1)

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integral from 0 to 5 of sin(-1+x)

\int_{0}^{5} sin (- 1 + x) d x

- cos (4) + cos (1)

Decimal

1.19394 \dots

Solution steps

\int_{0}^{5} sin (- 1 + x) d x

Apply Integration By Parts

: [sin (- 1 + x) x - \int cos (- 1 + x) x d x]_{0}^{5}

= [sin (- 1 + x) x - \int cos (- 1 + x) x d x]_{0}^{5}

\int cos (- 1 + x) x d x = - x sin (1 - x) + cos (1 - x)

= [sin (- 1 + x) x - (- x sin (1 - x) + cos (1 - x))]_{0}^{5}

Expand

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

= [x sin (- 1 + x) + x sin (1 - x) - cos (1 - x)]_{0}^{5}

Compute the boundaries:

- cos (4) + cos (1)

= - cos (4) + cos (1)

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What is the integral from 0 to 5 of sin(-1+x) ?
The integral from 0 to 5 of sin(-1+x) is -cos(4)+cos(1)