What is the integral of 8xcos(5x) ?

The integral of 8xcos(5x) is 8/25 (5xsin(5x)+cos(5x))+C

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integral of 8xcos(5x)

\int 8 x cos (5 x) d x

\frac{8}{25} (5 x sin (5 x) + cos (5 x)) + C

Solution steps

\int 8 x cos (5 x) d x

Take the constant out:

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

= 8 \cdot \int x cos (5 x) d x

Apply u-substitution

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

= 8 \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

= 8 \cdot \frac{1}{25} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 8 \cdot \frac{1}{25} (u sin (u) - \int sin (u) d u)

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

= 8 \cdot \frac{1}{25} (u sin (u) - (- cos (u)))

Substitute back

u = 5 x

= 8 \cdot \frac{1}{25} (5 x sin (5 x) - (- cos (5 x)))

Simplify

8 \cdot \frac{1}{25} (5 x sin (5 x) - (- cos (5 x))) : \frac{8}{25} (5 x sin (5 x) + cos (5 x))

= \frac{8}{25} (5 x sin (5 x) + cos (5 x))

Add a constant to the solution

= \frac{8}{25} (5 x sin (5 x) + cos (5 x)) + C

y^{^{''}} + 9 y^{^{'}} = 1

x \to - 2 lim (\frac{2 x}{x ^{2} - 4})

\int \frac{x ^{6} - x ^{3} + 25}{x ^{4} + 5 x ^{2}} d x

f (x) = e^{2} x

\frac{d}{d x} (sec (x^{3} e^{x}))

What is the integral of 8xcos(5x) ?
The integral of 8xcos(5x) is 8/25 (5xsin(5x)+cos(5x))+C