What is the integral from 0 to pi of 8sin(x/2) ?

The integral from 0 to pi of 8sin(x/2) is 16

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integral from 0 to pi of 8sin(x/2)

\int_{0}^{π} 8 sin (\frac{x}{2}) d x

16

Solution steps

\int_{0}^{π} 8 sin (\frac{x}{2}) d x

Take the constant out:

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

= 8 \cdot \int_{0}^{π} sin (\frac{x}{2}) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} sin (u) \cdot 2 d u

= 8 \cdot \int_{0}^{\frac{π}{2}} sin (u) \cdot 2 d u

Take the constant out:

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

= 8 \cdot 2 \cdot \int_{0}^{\frac{π}{2}} sin (u) d u

Use the common integral:

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

= 8 \cdot 2 [- cos (u)]_{0}^{\frac{π}{2}}

Simplify

= 16 [- cos (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= 16 \cdot 1

Simplify

= 16

y^{^{''}} - 6 y^{^{'}} + 8 y = - 2 cos (4 t)

\int_{0}^{2} 8 t d t

f (t) = t

\frac{d}{d x} (- \frac{2 x}{( x ^{2} - 4 ) ^{2}})

What is the integral from 0 to pi of 8sin(x/2) ?
The integral from 0 to pi of 8sin(x/2) is 16