What is the integral of 6x^2sin(4x) ?

The integral of 6x^2sin(4x) is 6(-1/4 x^2cos(4x)+1/32 (4xsin(4x)+cos(4x)))+C

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integral of 6x^2sin(4x)

\int 6 x^{2} sin (4 x) d x

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

Solution steps

\int 6 x^{2} sin (4 x) d x

Take the constant out:

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

= 6 \cdot \int x^{2} sin (4 x) d x

Apply Integration By Parts

: - \frac{1}{4} x^{2} cos (4 x) - \int - \frac{1}{2} x cos (4 x) d x

= 6 (- \frac{1}{4} x^{2} cos (4 x) - \int - \frac{1}{2} x cos (4 x) d x)

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

= 6 (- \frac{1}{4} x^{2} cos (4 x) - (- \frac{1}{32} (4 x sin (4 x) + cos (4 x))))

Simplify

= 6 (- \frac{1}{4} x^{2} cos (4 x) + \frac{1}{32} (4 x sin (4 x) + cos (4 x)))

Add a constant to the solution

= 6 (- \frac{1}{4} x^{2} cos (4 x) + \frac{1}{32} (4 x sin (4 x) + cos (4 x))) + C

\int \frac{1}{4 x ^{\frac{1}{2}}} d x

x^{2} y^{^{''}} - x (x + 2) y^{^{'}} + (x + 2) y = 0

x + 7 x - 7

x \to \infty lim (3 + x)

9 y^{^{''}} + 6 y^{^{'}} + y = e^{- \frac{1}{3} t}

What is the integral of 6x^2sin(4x) ?
The integral of 6x^2sin(4x) is 6(-1/4 x^2cos(4x)+1/32 (4xsin(4x)+cos(4x)))+C