What is the integral of q^2cos(7q) ?

The integral of q^2cos(7q) is 1/343 (49q^2sin(7q)-2(-7qcos(7q)+sin(7q)))+C

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integral of q^2cos(7q)

\int q^{2} cos (7 q) d q

\frac{1}{343} (49 q^{2} sin (7 q) - 2 (- 7 q cos (7 q) + sin (7 q))) + C

Solution steps

\int q^{2} cos (7 q) d q

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{343} d u

= \int \frac{u ^{2} cos ( u )}{343} d u

Take the constant out:

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

= \frac{1}{343} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

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

= \frac{1}{343} (u^{2} sin (u) - \int 2 u sin (u) d u)

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= \frac{1}{343} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 7 q

= \frac{1}{343} ((7 q)^{2} sin (7 q) - 2 (- 7 q cos (7 q) + sin (7 q)))

Simplify

\frac{1}{343} ((7 q)^{2} sin (7 q) - 2 (- 7 q cos (7 q) + sin (7 q))) : \frac{1}{343} (49 q^{2} sin (7 q) - 2 (- 7 q cos (7 q) + sin (7 q)))

= \frac{1}{343} (49 q^{2} sin (7 q) - 2 (- 7 q cos (7 q) + sin (7 q)))

Add a constant to the solution

= \frac{1}{343} (49 q^{2} sin (7 q) - 2 (- 7 q cos (7 q) + sin (7 q))) + C

\frac{d y}{d x} = \frac{( y - y ^{3} )}{a ^{2}}, y (0) = b

\frac{\partial}{\partial x} (b x e^{5 x y})

x \to (- 1) + lim (\frac{1}{x + 1})

x \to 2 lim (\frac{\frac{4}{x} - 2}{x - 2})

\frac{\partial}{\partial x} (z^{2})

What is the integral of q^2cos(7q) ?
The integral of q^2cos(7q) is 1/343 (49q^2sin(7q)-2(-7qcos(7q)+sin(7q)))+C