What is the integral from-3 to 3 of sin(x)x ?

The integral from-3 to 3 of sin(x)x is -6cos(3)+2sin(3)

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integral from-3 to 3 of sin(x)x

\int_{- 3}^{3} sin (x) x d x

- 6 cos (3) + 2 sin (3)

Decimal Notation

6.22219 \dots

Solution steps

\int_{- 3}^{3} sin (x) x d x

Apply Integration By Parts

: [- x cos (x) - \int - cos (x) d x]_{- 3}^{3}

= [- x cos (x) - \int - cos (x) d x]_{- 3}^{3}

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

= [- x cos (x) - (- sin (x))]_{- 3}^{3}

Simplify

= [- x cos (x) + sin (x)]_{- 3}^{3}

Compute the boundaries:

- 6 cos (3) + 2 sin (3)

= - 6 cos (3) + 2 sin (3)

\frac{d}{d x} (x^{\frac{9}{2}})

\int \frac{u}{1 - u ^{2}} d u

2 x d y - 2 y d x = x^{2} + 4 y^{2} d x

\int \frac{x ^{2}}{25 + x ^{6}} d x

\frac{\partial}{\partial x} (e^{- 4 t} cos (π x))

What is the integral from-3 to 3 of sin(x)x ?
The integral from-3 to 3 of sin(x)x is -6cos(3)+2sin(3)