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

The integral of x^3sin(x/2) is 16(-1/8 x^3cos(x/2)+3(1/4 x^2sin(x/2)+xcos(x/2)-2sin(x/2)))+C

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integral of x^3sin(x/2)

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

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

Solution steps

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

Apply u-substitution

: \int 16 u^{3} sin (u) d u

= \int 16 u^{3} sin (u) d u

Take the constant out:

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

= 16 \cdot \int u^{3} sin (u) d u

Apply Integration By Parts

: - u^{3} cos (u) - \int - 3 u^{2} cos (u) d u

= 16 (- u^{3} cos (u) - \int - 3 u^{2} cos (u) d u)

\int - 3 u^{2} cos (u) d u = - 3 (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

= 16 (- u^{3} cos (u) - (- 3 (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))))

Substitute back

u = \frac{x}{2}

= 16 (- (\frac{x}{2})^{3} cos (\frac{x}{2}) - (- 3 ((\frac{x}{2})^{2} sin (\frac{x}{2}) - 2 (- \frac{x}{2} cos (\frac{x}{2}) + sin (\frac{x}{2})))))

Simplify

16 (- (\frac{x}{2})^{3} cos (\frac{x}{2}) - (- 3 ((\frac{x}{2})^{2} sin (\frac{x}{2}) - 2 (- \frac{x}{2} cos (\frac{x}{2}) + sin (\frac{x}{2}))))) : 16 (- \frac{1}{8} x^{3} cos (\frac{x}{2}) + 3 (\frac{1}{4} x^{2} sin (\frac{x}{2}) + x cos (\frac{x}{2}) - 2 sin (\frac{x}{2})))

= 16 (- \frac{1}{8} x^{3} cos (\frac{x}{2}) + 3 (\frac{1}{4} x^{2} sin (\frac{x}{2}) + x cos (\frac{x}{2}) - 2 sin (\frac{x}{2})))

Add a constant to the solution

= 16 (- \frac{1}{8} x^{3} cos (\frac{x}{2}) + 3 (\frac{1}{4} x^{2} sin (\frac{x}{2}) + x cos (\frac{x}{2}) - 2 sin (\frac{x}{2}))) + C

\int \frac{1 + cos ( 2 t )}{2} d t

\int \frac{6 x ^{2} + x}{( 16 x ^{3} + 4 x ^{2} ) ^{2}} d x

\int \frac{2}{1 + t ^{2}} d t

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

\int \frac{x ^{3} + x ^{2} + 2}{x ^{4} + 4 x ^{2} + 4} d x

What is the integral of x^3sin(x/2) ?
The integral of x^3sin(x/2) is 16(-1/8 x^3cos(x/2)+3(1/4 x^2sin(x/2)+xcos(x/2)-2sin(x/2)))+C