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

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

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integral of 8x^2cos(2x)

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

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

Solution steps

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

Apply Integration By Parts

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

= 8 (\frac{1}{2} x^{2} sin (2 x) - \int x sin (2 x) d x)

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

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

Simplify

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

Add a constant to the solution

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

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

n = 1 \sum \infty (\frac{1}{n}) (\frac{1}{n})^{2}

y^{^{'}} = \frac{2 x y}{x ^{2} + 1}

f (x) = 6 + 1 - x, at x = - 8

\int \frac{x - 8}{( x + 4 ) ^{2} + 64} d x

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