What is the integral of 5xcos(6x) ?

The integral of 5xcos(6x) is 5/36 (6xsin(6x)+cos(6x))+C

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integral of 5xcos(6x)

\int 5 x cos (6 x) d x

\frac{5}{36} (6 x sin (6 x) + cos (6 x)) + C

Solution steps

\int 5 x cos (6 x) d x

Take the constant out:

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

= 5 \cdot \int x cos (6 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{36} d u

= 5 \cdot \int \frac{u cos ( u )}{36} d u

Take the constant out:

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

= 5 \cdot \frac{1}{36} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 5 \cdot \frac{1}{36} (u sin (u) - \int sin (u) d u)

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

= 5 \cdot \frac{1}{36} (u sin (u) - (- cos (u)))

Substitute back

u = 6 x

= 5 \cdot \frac{1}{36} (6 x sin (6 x) - (- cos (6 x)))

Simplify

5 \cdot \frac{1}{36} (6 x sin (6 x) - (- cos (6 x))) : \frac{5}{36} (6 x sin (6 x) + cos (6 x))

= \frac{5}{36} (6 x sin (6 x) + cos (6 x))

Add a constant to the solution

= \frac{5}{36} (6 x sin (6 x) + cos (6 x)) + C

x \to 0 lim ((1 + \frac{1}{x})^{3 x})

x \frac{d y}{d x} - 4 y = 6 x^{4} e^{x}

\int_{0}^{5} (22 + t) d t

- \frac{4}{( 4 x - 3 ) ^{\frac{3}{2}}}

\frac{ln ( 4 x )}{x}

What is the integral of 5xcos(6x) ?
The integral of 5xcos(6x) is 5/36 (6xsin(6x)+cos(6x))+C