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

The integral of x^3*sin^2((pix)/L) is

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integral of x^3*sin^2((pix)/L)

\int x^{3} \cdot sin^{2} (\frac{π x}{L})

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

Solution steps

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

Apply u-substitution

: \int \frac{L ^{4} u ^{3} sin ^{2} ( u )}{π ^{4}} d u

= \int \frac{L ^{4} u ^{3} sin ^{2} ( u )}{π ^{4}} d u

Take the constant out:

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

= \frac{L ^{4}}{π ^{4}} \cdot \int u^{3} sin^{2} (u) d u

Apply Integration By Parts

: \frac{1}{4} u^{4} sin^{2} (u) - \int \frac{1}{4} u^{4} sin (2 u) d u

= \frac{L ^{4}}{π ^{4}} (\frac{1}{4} u^{4} sin^{2} (u) - \int \frac{1}{4} u^{4} sin (2 u) d u)

\int \frac{1}{4} u^{4} sin (2 u) d u = \frac{1}{16} (- 2 u^{4} cos (2 u) + 4 u^{3} sin (2 u) - 3 (- 2 u^{2} cos (2 u) + 2 u sin (2 u) + cos (2 u)))

= \frac{L ^{4}}{π ^{4}} (\frac{1}{4} u^{4} sin^{2} (u) - \frac{1}{16} (- 2 u^{4} cos (2 u) + 4 u^{3} sin (2 u) - 3 (- 2 u^{2} cos (2 u) + 2 u sin (2 u) + cos (2 u))))

Substitute back

u = \frac{x π}{L}

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

Simplify

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

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

Add a constant to the solution

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

6 sin (2 x)

\frac{d}{d x} (1 - \frac{4}{x + 3})

n = 0 \sum \infty (\frac{e}{π})^{n}

\int cot (3 x) csc^{4} (3 x) d x

\int 45

What is the integral of x^3*sin^2((pix)/L) ?
The integral of x^3*sin^2((pix)/L) is