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

The integral of 9xcos(2x) is 9/4 (2xsin(2x)+cos(2x))+C

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integral of 9xcos(2x)

\int 9 x cos (2 x) d x

\frac{9}{4} (2 x sin (2 x) + cos (2 x)) + C

Solution steps

\int 9 x cos (2 x) d x

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u cos ( u )}{4} d u

= 9 \cdot \int \frac{u cos ( u )}{4} d u

Take the constant out:

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

= 9 \cdot \frac{1}{4} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 9 \cdot \frac{1}{4} (u sin (u) - \int sin (u) d u)

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

= 9 \cdot \frac{1}{4} (u sin (u) - (- cos (u)))

Substitute back

u = 2 x

= 9 \cdot \frac{1}{4} (2 x sin (2 x) - (- cos (2 x)))

Simplify

9 \cdot \frac{1}{4} (2 x sin (2 x) - (- cos (2 x))) : \frac{9}{4} (2 x sin (2 x) + cos (2 x))

= \frac{9}{4} (2 x sin (2 x) + cos (2 x))

Add a constant to the solution

= \frac{9}{4} (2 x sin (2 x) + cos (2 x)) + C

y = x^{2} - 1, (- 1, 1)

y = x - 6, y^{2} = x

tan^{2} (5 x)

\int_{1}^{2} x 7^{x^{2}} d x

\int t^{2} + 2 t + 1 d t

What is the integral of 9xcos(2x) ?
The integral of 9xcos(2x) is 9/4 (2xsin(2x)+cos(2x))+C