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

The integral of 5xcos(5x) is xsin(5x)+1/5 cos(5x)+C

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integral of 5xcos(5x)

\int 5 x cos (5 x) d x

x sin (5 x) + \frac{1}{5} cos (5 x) + C

Solution steps

\int 5 x cos (5 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 (5 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{25} d u

= 5 \cdot \int \frac{u cos ( u )}{25} d u

Take the constant out:

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

= 5 \cdot \frac{1}{25} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 5 \cdot \frac{1}{25} (u sin (u) - \int sin (u) d u)

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

= 5 \cdot \frac{1}{25} (u sin (u) - (- cos (u)))

Substitute back

u = 5 x

= 5 \cdot \frac{1}{25} (5 x sin (5 x) - (- cos (5 x)))

Simplify

5 \cdot \frac{1}{25} (5 x sin (5 x) - (- cos (5 x))) : x sin (5 x) + \frac{1}{5} cos (5 x)

= x sin (5 x) + \frac{1}{5} cos (5 x)

Add a constant to the solution

= x sin (5 x) + \frac{1}{5} cos (5 x) + C

f (x) = x^{2} - 5 x + 3

\int 4 sin (y) d y

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\frac{d y}{d x} = \frac{1 - x y ^{2}}{2 x ^{2} y}

\frac{\partial}{\partial x} (z (2 x y))

What is the integral of 5xcos(5x) ?
The integral of 5xcos(5x) is xsin(5x)+1/5 cos(5x)+C