What is the integral of bx*sin((pix)/(2a)) ?

The integral of bx*sin((pix)/(2a)) is b(-(2a)/pi xcos((pix)/(2a))+(4a^2)/(pi^2)sin((pix)/(2a)))+C

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integral of bx*sin((pix)/(2a))

\int b x \cdot sin (\frac{π x}{2 a}) d x

b (- \frac{2 a}{π} x cos (\frac{π x}{2 a}) + \frac{4 a ^{2}}{π ^{2}} sin (\frac{π x}{2 a})) + C

Solution steps

\int b x sin (\frac{π x}{2 a}) d x

Take the constant out:

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

= b \cdot \int x sin (\frac{π x}{2 a}) d x

Apply Integration By Parts

: - \frac{2 a}{π} x cos (\frac{π x}{2 a}) - \int - \frac{2 a}{π} cos (\frac{π x}{2 a}) d x

= b (- \frac{2 a}{π} x cos (\frac{π x}{2 a}) - \int - \frac{2 a}{π} cos (\frac{π x}{2 a}) d x)

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

= b (- \frac{2 a}{π} x cos (\frac{π x}{2 a}) - (- \frac{4 a ^{2}}{π ^{2}} sin (\frac{π x}{2 a})))

Simplify

= b (- \frac{2 a}{π} x cos (\frac{π x}{2 a}) + \frac{4 a ^{2}}{π ^{2}} sin (\frac{π x}{2 a}))

Add a constant to the solution

= b (- \frac{2 a}{π} x cos (\frac{π x}{2 a}) + \frac{4 a ^{2}}{π ^{2}} sin (\frac{π x}{2 a})) + C

\frac{\partial}{\partial y} (5 x^{2} - 2 x y + 3 y^{3})

(5.8) (3.9)

y^{^{''}} - y = \frac{1}{e ^{x} + e ^{- x}}

\frac{\partial}{\partial x} (cos (x y))

\frac{d}{d x} (\frac{tan ( x )}{1 + cos ( x )})

What is the integral of bx*sin((pix)/(2a)) ?
The integral of bx*sin((pix)/(2a)) is b(-(2a)/pi xcos((pix)/(2a))+(4a^2)/(pi^2)sin((pix)/(2a)))+C