What is the integral of-3/2 x^2cos(2x) ?

The integral of-3/2 x^2cos(2x) is -3/8 (2x^2sin(2x)+2xcos(2x)-sin(2x))+C

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integral of-3/2 x^2cos(2x)

\int - \frac{3}{2} x^{2} cos (2 x) d x

- \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)) + C

Solution steps

\int - \frac{3}{2} 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

= - \frac{3}{2} \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

= - \frac{3}{2} (\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)

= - \frac{3}{2} (\frac{1}{2} x^{2} sin (2 x) - (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)))

Simplify

- \frac{3}{2} (\frac{1}{2} x^{2} sin (2 x) - (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x))) : - \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

= - \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

Add a constant to the solution

= - \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)) + C

\int \frac{x}{1 - 4 x} d x

\int - \frac{sin ( 3 x )}{cos ^{3} ( 3 x )} d x

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

\int \frac{x ^{2} - 5}{( 1 + x ) ^{3}} d x

\frac{1}{4 x}

What is the integral of-3/2 x^2cos(2x) ?
The integral of-3/2 x^2cos(2x) is -3/8 (2x^2sin(2x)+2xcos(2x)-sin(2x))+C