What is the integral of 8x^2cos(5x) ?

The integral of 8x^2cos(5x) is 8/125 (25x^2sin(5x)-2(-5xcos(5x)+sin(5x)))+C

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integral of 8x^2cos(5x)

\int 8 x^{2} cos (5 x) d x

\frac{8}{125} (25 x^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x))) + C

Solution steps

\int 8 x^{2} 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^{2} cos (5 x) d x

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{125} d u

= 8 \cdot \int \frac{u ^{2} cos ( u )}{125} d u

Take the constant out:

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

= 8 \cdot \frac{1}{125} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

: u^{2} sin (u) - \int 2 u sin (u) d u

= 8 \cdot \frac{1}{125} (u^{2} sin (u) - \int 2 u sin (u) d u)

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= 8 \cdot \frac{1}{125} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 5 x

= 8 \cdot \frac{1}{125} ((5 x)^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x)))

Simplify

8 \cdot \frac{1}{125} ((5 x)^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x))) : \frac{8}{125} (25 x^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x)))

= \frac{8}{125} (25 x^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x)))

Add a constant to the solution

= \frac{8}{125} (25 x^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x))) + C

f (x) = \frac{x}{x ^{2} + 10}

\int 10 sin (x) d x

x \to 2 + lim (\frac{x}{∣ x ∣})

(- 45) (- 4 - 8)

f (x) = - \frac{3}{e ^{3 x}}, x = 0

What is the integral of 8x^2cos(5x) ?
The integral of 8x^2cos(5x) is 8/125 (25x^2sin(5x)-2(-5xcos(5x)+sin(5x)))+C