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

The integral from 0 to pi/2 of 8cos(x)sin(sin(x)) is 8(-cos(1)+1)

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

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

8 (- cos (1) + 1)

Decimal Notation

3.67758 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{1} sin (u) d u

= 8 \cdot \int_{0}^{1} sin (u) d u

Use the common integral:

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

= 8 [- cos (u)]_{0}^{1}

Compute the boundaries:

- cos (1) + 1

= 8 (- cos (1) + 1)

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What is the integral from 0 to pi/2 of 8cos(x)sin(sin(x)) ?
The integral from 0 to pi/2 of 8cos(x)sin(sin(x)) is 8(-cos(1)+1)