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

The integral of 2x^2cos(4x) is 1/16 (8x^2sin(4x)-sin(4x)+4xcos(4x))+C

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integral of 2x^2cos(4x)

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

\frac{1}{16} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x)) + C

Solution steps

\int 2 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

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

Apply u-substitution

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

= 2 \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

= 2 \cdot \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

= 2 \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))

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

Substitute back

u = 4 x

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

Simplify

2 \cdot \frac{1}{64} ((4 x)^{2} sin (4 x) - 2 (- 4 x cos (4 x) + sin (4 x))) : \frac{1}{16} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x))

= \frac{1}{16} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x))

Add a constant to the solution

= \frac{1}{16} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x)) + C

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\int \frac{3 x - 5}{- x ^{5}} d x

\int \frac{1}{( a ^{2} + x ^{2} ) ^{3}} d x

What is the integral of 2x^2cos(4x) ?
The integral of 2x^2cos(4x) is 1/16 (8x^2sin(4x)-sin(4x)+4xcos(4x))+C