What is the integral of 2xcos(9x) ?

The integral of 2xcos(9x) is 2/81 (9xsin(9x)+cos(9x))+C

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integral of 2xcos(9x)

\int 2 x cos (9 x) d x

\frac{2}{81} (9 x sin (9 x) + cos (9 x)) + C

Solution steps

\int 2 x cos (9 x) d x

Take the constant out:

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

= 2 \cdot \int x cos (9 x) d x

Apply u-substitution

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

= 2 \cdot \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

= 2 \cdot \frac{1}{81} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{81} (u sin (u) - \int sin (u) d u)

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

= 2 \cdot \frac{1}{81} (u sin (u) - (- cos (u)))

Substitute back

u = 9 x

= 2 \cdot \frac{1}{81} (9 x sin (9 x) - (- cos (9 x)))

Simplify

2 \cdot \frac{1}{81} (9 x sin (9 x) - (- cos (9 x))) : \frac{2}{81} (9 x sin (9 x) + cos (9 x))

= \frac{2}{81} (9 x sin (9 x) + cos (9 x))

Add a constant to the solution

= \frac{2}{81} (9 x sin (9 x) + cos (9 x)) + C

f (x) = \frac{3}{x - 2}, at x = 3

\int_{0}^{4} \frac{3 t}{( t - 5 ) ^{2}} d t

f (x) = \frac{x}{x ^{2} + 6}

x \to 2 lim (5 x^{3} - 9 x^{2} + 1)

x \to \infty lim (4^{x} x)

What is the integral of 2xcos(9x) ?
The integral of 2xcos(9x) is 2/81 (9xsin(9x)+cos(9x))+C