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

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

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

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

- \frac{1}{4} (2 x sin (2 x) + cos (2 x)) + C

Solution steps

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

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

= - 1 \cdot \int cos (2 x) x d x

Apply u-substitution

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

= - 1 \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 2 x

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

Simplify

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

Add a constant to the solution

= - \frac{1}{4} (2 x sin (2 x) + cos (2 x)) + C

\frac{d}{d x} (\frac{x + 3}{7 x ^{2} e ^{x}})

\frac{d}{d x} (16 cos (x))

\int π \cdot x^{2} d x

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9 x^{2} x + 2 x^{2}

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