What is the integral of-3xcos(4x) ?

The integral of-3xcos(4x) is -3/16 (4xsin(4x)+cos(4x))+C

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integral of-3xcos(4x)

\int - 3 x cos (4 x) d x

- \frac{3}{16} (4 x sin (4 x) + cos (4 x)) + C

Solution steps

\int - 3 x cos (4 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 (4 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{16} d u

= - 3 \cdot \int \frac{u cos ( u )}{16} d u

Take the constant out:

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

= - 3 \cdot \frac{1}{16} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= - 3 \cdot \frac{1}{16} (u sin (u) - \int sin (u) d u)

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

= - 3 \cdot \frac{1}{16} (u sin (u) - (- cos (u)))

Substitute back

u = 4 x

= - 3 \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x)))

Simplify

- 3 \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x))) : - \frac{3}{16} (4 x sin (4 x) + cos (4 x))

= - \frac{3}{16} (4 x sin (4 x) + cos (4 x))

Add a constant to the solution

= - \frac{3}{16} (4 x sin (4 x) + cos (4 x)) + C

\int (- 5 x^{4} + 8 x^{3}) (- 2 x^{5} + 4 x^{4})^{6} d x

f (x) = 3 x^{2} + 6

\int - x^{2} - 1 d x

y^{^{'}} = \frac{x - y + 1}{x + y - 3}

f (x) = \frac{2 - ln ( x ^{2} )}{x ^{2}}

What is the integral of-3xcos(4x) ?
The integral of-3xcos(4x) is -3/16 (4xsin(4x)+cos(4x))+C