Indefinite Integral Inverse Trigonometric Function

Question

$$\int \frac{2}{9w^2+25}dw$$ I already know this will be equal to $\frac{1}{a} \arctan(x/a)$, but I don't know how to factor out the $9$. I only know how to take out the $2$.

Answer 1

Well you can easily reduce the integrand to $2/9 \cdot \frac{1}{w^2+(5/3)^2}$ or possibly $2/25 \cdot \frac{1}{(3w/5)^2+1}$. What does this tell you?

Answer 2

$$ \begin{align} & \phantom{={}} 2\int \frac{dw}{9w^2+25} = 2\int\frac{dw}{25(\frac{9}{25} w^2 + 1)} = \frac{2}{25}\int\frac{dw}{\left(\frac35 w\right)^2 + 1} \\[12pt] & = \frac{2}{3\cdot5} \int \frac{\frac35\, dw}{\left(\frac35 w\right)^2 + 1} = \frac{2}{15} \int \frac{du}{u^2+1}\text{ etc.} \end{align} $$