What is the integral of 4x^3sin(2x) ?

The integral of 4x^3sin(2x) is 4(-1/2 x^3cos(2x)+3/8 (2x^2sin(2x)+2xcos(2x)-sin(2x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4x^3sin(2x)

\int 4 x^{3} sin (2 x) d x

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

Solution steps

\int 4 x^{3} sin (2 x) d x

Take the constant out:

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

= 4 \cdot \int x^{3} sin (2 x) d x

Apply Integration By Parts

: - \frac{1}{2} x^{3} cos (2 x) - \int - \frac{3}{2} x^{2} cos (2 x) d x

= 4 (- \frac{1}{2} x^{3} cos (2 x) - \int - \frac{3}{2} x^{2} cos (2 x) d x)

\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))

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

Simplify

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

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

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

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

\int d (x^{3}) d x

\int cos^{2} (x) sin^{1} (x) d x

\int \frac{x}{x ^{2} - 6 x + 8} d x

What is the integral of 4x^3sin(2x) ?
The integral of 4x^3sin(2x) is 4(-1/2 x^3cos(2x)+3/8 (2x^2sin(2x)+2xcos(2x)-sin(2x)))+C