Evaluating $\;\int\sqrt{2uv}~dv\;?$

Question

How do you evaluate $\quad\displaystyle \int\sqrt{2uv}\,dv\;?$

I tried for half an hour using integration of parts and other "methods" but I can't seem to get the answer. I think it's the square root that is confusing me.. as well as the fact that there's a second variable (which I would treat as a constant).

Can someone please help me? Thanks in advance.

Answer 1

You should treat $u$ as a constant (the "$dv$" tells you to integrate strictly in terms of $v$), so we consider all of $\sqrt{2u}$ to be a "constant."

$$\int\sqrt{2uv}~dv = \quad \int \sqrt{2u}\cdot \sqrt v = \quad \sqrt{2u}\int \sqrt v \,dv = \quad \sqrt{2u}\int v^{1/2} dv$$

Can you take it from here?

Remember the rule for integrating with respect to a variable raised to a power:

$$\int v^a \,dv = \dfrac{v^{a + 1}}{a+1} + \;\text{Constant}\quad \text{if} \;\;a\neq -1$$

Answer 2

$$\int\sqrt{2uv}dv=\sqrt2u\int v^{1/2}dv=\sqrt{2u}\frac{v^{3/2}}{3/2}+K\ldots$$