What is the integral from 0 to sqrt(x of)cos(9t) ?

The integral from 0 to sqrt(x of)cos(9t) is 1/9 sin(9sqrt(x))

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integral from 0 to sqrt(x of)cos(9t)

\int_{0}^{x} cos (9 t) d t

\frac{1}{9} sin (9 x)

Solution steps

\int_{0}^{x} cos (9 t) d t

Apply u-substitution

: \int_{0}^{9 x} cos (u) \frac{1}{9} d u

= \int_{0}^{9 x} cos (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{0}^{9 x} cos (u) d u

Use the common integral:

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

= \frac{1}{9} [sin (u)]_{0}^{9 x}

Compute the boundaries:

sin (9 x)

= \frac{1}{9} sin (9 x)

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What is the integral from 0 to sqrt(x of)cos(9t) ?
The integral from 0 to sqrt(x of)cos(9t) is 1/9 sin(9sqrt(x))