What is the integral of xcos(9x) ?

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

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integral of xcos(9x)

\int x cos (9 x) d x

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

Solution steps

\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 6 lim (9)

\frac{d y}{d t} + t^{2} y = 1

\frac{\partial}{\partial x} (e^{x} cos (y))

y^{^{'}} - \frac{2}{x ^{2}} y = 0

f (x) = 2 x^{2} ln (2 x)

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