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

The integral of x^2cos(4x) is 1/32 (8x^2sin(4x)-sin(4x)+4xcos(4x))+C

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integral of x^2cos(4x)

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

\frac{1}{32} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x)) + C

Solution steps

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

Apply u-substitution

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

= \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

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

Apply Integration By Parts

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

= \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))

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

Substitute back

u = 4 x

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

Simplify

\frac{1}{64} ((4 x)^{2} sin (4 x) - 2 (- 4 x cos (4 x) + sin (4 x))) : \frac{1}{32} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x))

= \frac{1}{32} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x))

Add a constant to the solution

= \frac{1}{32} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x)) + C

\int x + y d x

f (x) = sin (8 e^{x})

\int (3 x^{2} y) d x

sin^{2} (x), sin^{3} (x), [0, 2 π]

y^{^{'}} = \frac{x ^{8} - 8 y}{x}

What is the integral of x^2cos(4x) ?
The integral of x^2cos(4x) is 1/32 (8x^2sin(4x)-sin(4x)+4xcos(4x))+C