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

The integral from-1 to pi-1 of sin(1+x) is 2

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integral from-1 to pi-1 of sin(1+x)

\int_{- 1}^{π - 1} sin (1 + x) d x

2

Solution steps

\int_{- 1}^{π - 1} sin (1 + x) d x

Apply u-substitution

: \int_{0}^{π} sin (u) d u

= \int_{0}^{π} sin (u) d u

Use the common integral:

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

= [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= 2

\int_{0}^{2} x^{2} + 2 d x

\int_{0}^{z^{2}} z d z

\int_{0}^{4} (4 - 4 x) d x

\int_{0}^{1} x \cdot 6 x (1 - x) d x

\int_{5}^{11} \frac{y}{y ^{2} - y - 2} d y

What is the integral from-1 to pi-1 of sin(1+x) ?
The integral from-1 to pi-1 of sin(1+x) is 2