What is the integral of 3xcos(8x) ?

The integral of 3xcos(8x) is 3/64 (8xsin(8x)+cos(8x))+C

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integral of 3xcos(8x)

\int 3 x cos (8 x) d x

\frac{3}{64} (8 x sin (8 x) + cos (8 x)) + C

Solution steps

\int 3 x cos (8 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 (8 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{64} d u

= 3 \cdot \int \frac{u cos ( u )}{64} d u

Take the constant out:

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

= 3 \cdot \frac{1}{64} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{64} (u sin (u) - \int sin (u) d u)

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

= 3 \cdot \frac{1}{64} (u sin (u) - (- cos (u)))

Substitute back

u = 8 x

= 3 \cdot \frac{1}{64} (8 x sin (8 x) - (- cos (8 x)))

Simplify

3 \cdot \frac{1}{64} (8 x sin (8 x) - (- cos (8 x))) : \frac{3}{64} (8 x sin (8 x) + cos (8 x))

= \frac{3}{64} (8 x sin (8 x) + cos (8 x))

Add a constant to the solution

= \frac{3}{64} (8 x sin (8 x) + cos (8 x)) + C

n = 1 \sum \infty 5 cos (\frac{2}{n})

6 x^{2} sin (x) tan (x)

\frac{d y}{d x}, x = y^{2}

\frac{d}{d x} (\frac{400}{x})

\frac{d y}{d x} = \frac{x ^{2} - 7}{15 y ^{2}}

What is the integral of 3xcos(8x) ?
The integral of 3xcos(8x) is 3/64 (8xsin(8x)+cos(8x))+C