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

The integral of (x^2-2)cos(2x) is 1/2 x^2sin(2x)+1/2 xcos(2x)-5/4 sin(2x)+C

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integral of (x^2-2)cos(2x)

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

\frac{1}{2} x^{2} sin (2 x) + \frac{1}{2} x cos (2 x) - \frac{5}{4} sin (2 x) + C

Solution steps

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

Expand

(x^{2} - 2) cos (2 x) : x^{2} cos (2 x) - 2 cos (2 x)

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 2 cos (2 x) d x = sin (2 x)

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

Add similar elements:

- \frac{1}{4} sin (2 x) - sin (2 x) = - \frac{5}{4} sin (2 x)

= \frac{1}{2} x^{2} sin (2 x) + \frac{1}{2} x cos (2 x) - \frac{5}{4} sin (2 x)

Add a constant to the solution

= \frac{1}{2} x^{2} sin (2 x) + \frac{1}{2} x cos (2 x) - \frac{5}{4} sin (2 x) + C

\int (sin (x) sin (3 x)) d x

\int \frac{x ^{3} + 2 x ^{2} + 3 x + 7}{x ^{2} - x - 6} d x

\int \frac{g}{k} (1 - e^{- k t}) d t

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

What is the integral of (x^2-2)cos(2x) ?
The integral of (x^2-2)cos(2x) is 1/2 x^2sin(2x)+1/2 xcos(2x)-5/4 sin(2x)+C