What is the integral of 64x^2cos(4x) ?

The integral of 64x^2cos(4x) is 16x^2sin(4x)-2(-4xcos(4x)+sin(4x))+C

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integral of 64x^2cos(4x)

\int 64 x^{2} cos (4 x) d x

16 x^{2} sin (4 x) - 2 (- 4 x cos (4 x) + sin (4 x)) + C

Solution steps

\int 64 x^{2} cos (4 x) d x

Take the constant out:

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

= 64 \cdot \int x^{2} cos (4 x) d x

Apply u-substitution

= 64 \cdot \int \frac{u ^{2} cos ( u )}{64} d u

Take the constant out:

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

= 64 \cdot \frac{1}{64} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

= 64 \cdot \frac{1}{64} (u^{2} sin (u) - \int 2 u sin (u) d u)

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

= 64 \cdot \frac{1}{64} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 4 x

= 64 \cdot \frac{1}{64} ((4 x)^{2} sin (4 x) - 2 (- 4 x cos (4 x) + sin (4 x)))

Simplify

64 \cdot \frac{1}{64} ((4 x)^{2} sin (4 x) - 2 (- 4 x cos (4 x) + sin (4 x))) : 16 x^{2} sin (4 x) - 2 (- 4 x cos (4 x) + sin (4 x))

= 16 x^{2} sin (4 x) - 2 (- 4 x cos (4 x) + sin (4 x))

Add a constant to the solution

= 16 x^{2} sin (4 x) - 2 (- 4 x cos (4 x) + sin (4 x)) + C

d er i v a t i v e g (t) = \frac{9}{t}

\frac{d y}{d x} + 6 x^{2} y = e^{- 2 x^{3}}

\frac{d}{d x} (arccsc (2 x))

\frac{d y}{d t} - 3 y = 3 e^{4 t}

d er i v a t i v e y = 2 x^{2} lo g_{e} (x) - x^{2}

What is the integral of 64x^2cos(4x) ?
The integral of 64x^2cos(4x) is 16x^2sin(4x)-2(-4xcos(4x)+sin(4x))+C