What is the integral of-1/4 cos(4x)2x ?

The integral of-1/4 cos(4x)2x is -1/32 (4xsin(4x)+cos(4x))+C

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integral of-1/4 cos(4x)2x

\int - \frac{1}{4} cos (4 x) 2 x d x

- \frac{1}{32} (4 x sin (4 x) + cos (4 x)) + C

Solution steps

\int - \frac{1}{4} cos (4 x) \cdot 2 x d x

Take the constant out:

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

= - \frac{1}{4} \cdot 2 \cdot \int cos (4 x) x d x

- \frac{1}{4} \cdot 2 = - \frac{1}{2}

= - \frac{1}{2} \cdot \int cos (4 x) x d x

Apply u-substitution

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

= - \frac{1}{2} \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

= - \frac{1}{2} \cdot \frac{1}{16} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= - \frac{1}{2} \cdot \frac{1}{16} (u sin (u) - \int sin (u) d u)

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

= - \frac{1}{2} \cdot \frac{1}{16} (u sin (u) - (- cos (u)))

Substitute back

u = 4 x

= - \frac{1}{2} \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x)))

Simplify

- \frac{1}{2} \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x))) : - \frac{1}{32} (4 x sin (4 x) + cos (4 x))

= - \frac{1}{32} (4 x sin (4 x) + cos (4 x))

Add a constant to the solution

= - \frac{1}{32} (4 x sin (4 x) + cos (4 x)) + C

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What is the integral of-1/4 cos(4x)2x ?
The integral of-1/4 cos(4x)2x is -1/32 (4xsin(4x)+cos(4x))+C