What is the integral of (-3x)cos(2x) ?

The integral of (-3x)cos(2x) is -3/4 (2xsin(2x)+cos(2x))+C

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integral of (-3x)cos(2x)

\int (- 3 x) cos (2 x) d x

- \frac{3}{4} (2 x sin (2 x) + cos (2 x)) + C

Solution steps

\int (- 3 x) cos (2 x) d x

Rewrite using trig identities

= \int - 3 x (cos^{2} (x) - sin^{2} (x)) d x

Take the constant out:

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

= - 3 \cdot \int x (cos^{2} (x) - sin^{2} (x)) d x

Rewrite using trig identities

= - 3 \cdot \int x cos (2 x) d x

Apply u-substitution

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

= - 3 \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

= - 3 \cdot \frac{1}{4} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= - 3 \cdot \frac{1}{4} (u sin (u) - \int sin (u) d u)

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

= - 3 \cdot \frac{1}{4} (u sin (u) - (- cos (u)))

Substitute back

u = 2 x

= - 3 \cdot \frac{1}{4} (2 x sin (2 x) - (- cos (2 x)))

Simplify

- 3 \cdot \frac{1}{4} (2 x sin (2 x) - (- cos (2 x))) : - \frac{3}{4} (2 x sin (2 x) + cos (2 x))

= - \frac{3}{4} (2 x sin (2 x) + cos (2 x))

Add a constant to the solution

= - \frac{3}{4} (2 x sin (2 x) + cos (2 x)) + C

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What is the integral of (-3x)cos(2x) ?
The integral of (-3x)cos(2x) is -3/4 (2xsin(2x)+cos(2x))+C