Integral dx/sqrt(x - a)integral dx/sqrt(1/ax)

[rebeka]

integral dx/sqrt(x - a)
integral dx/sqrt(1/ax)

 

[rebeka]

reading suggestions

 

[rebeka]

sorry for wasting time I think I just understood substitution, out of embarassment I am never coming back to this forum(bright side of life I learned more in three days than I have in two years)

 

[HallsofIvy]

We all embarrass ourselves at one time or another! (Some of us more than others. Believe me, I know about embarrassing myself!) PLEASE come back to PhysicsForums!

(I need someone to make me look good!)

Anyway, for those who are still wondering how to do these:

$$\int \sqrt{x-a}dx$$

Let u= x-1 so du= dx and $$\sqrt{x-a}= \sqrt{u}= u^{\frac{1}{2}}$$. The integral becomes $$\int u^{\frac{1}{2}}du$$ which can be done with the power law.

$$\int \frac{dx}{\sqrt{\frac{1}{ax}}}$$

$$\frac{1}{\sqrt{\frac{1}{ax}}}$$ is just $$\sqrt{ax}$$.

Let u= ax so du= adx, dx= (1/a)du and $$\sqrt{ax}= \sqrt{u}= u^{\frac{1}{2}}$$

The integral becomes $$\frac{1}{a}\int u^{\frac{1}{2}}du$$

 

[Orion1]

$$\int \sqrt{x-a}dx$$

HallsofIvy, should not this equation be:
$$\int \frac{dx}{\sqrt{x - a}} = \int u^{-\frac{1}{2}} du$$

 

[HallsofIvy]

Oh, you're right! I missed the "1/" in the first post. Of course, the substitution would be exactly the same and the u-integral what you show. Thanks.