What is the integral of x^1cos(-9x) ?

The integral of x^1cos(-9x) is 1/81 (9xsin(9x)+cos(9x))+C

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integral of x^1cos(-9x)

\int x^{1} cos (- 9 x) d x

\frac{1}{81} (9 x sin (9 x) + cos (9 x)) + C

Solution steps

\int x^{1} cos (- 9 x) d x

Simplify

x^{1} cos (- 9 x) : x cos (9 x)

= \int x cos (9 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{81} d u

= \int \frac{u cos ( u )}{81} d u

Take the constant out:

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

= \frac{1}{81} \cdot \int u cos (u) d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

= \frac{1}{81} (u sin (u) - \int sin (u) d u)

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

= \frac{1}{81} (u sin (u) - (- cos (u)))

Substitute back

u = 9 x

= \frac{1}{81} (9 x sin (9 x) - (- cos (9 x)))

Simplify

= \frac{1}{81} (9 x sin (9 x) + cos (9 x))

Add a constant to the solution

= \frac{1}{81} (9 x sin (9 x) + cos (9 x)) + C

x \to 3 lim (\frac{∣ x - 2 ∣}{x - 2})

y = \frac{4 x ^{7}}{7}

\frac{d}{d x} (e^{3 x - 5})

\int arcsin (x) d x

3 x - 2

What is the integral of x^1cos(-9x) ?
The integral of x^1cos(-9x) is 1/81 (9xsin(9x)+cos(9x))+C