sorry for wasting time I think I just understood substitution, out of embarassment im never coming back to this forum(bright side of life I learned more in three days than I have in two years)
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:
[tex]\int \sqrt{x-a}dx[/tex]
Let u= x-1 so du= dx and [tex]\sqrt{x-a}= \sqrt{u}= u^{\frac{1}{2}}[/tex]. The integral becomes [tex]\int u^{\frac{1}{2}}du[/tex] which can be done with the power law.
[tex]\int \frac{dx}{\sqrt{\frac{1}{ax}}}[/tex]
[tex]\frac{1}{\sqrt{\frac{1}{ax}}}[/tex] is just [tex]\sqrt{ax}[/tex].
Let u= ax so du= adx, dx= (1/a)du and [tex]\sqrt{ax}= \sqrt{u}= u^{\frac{1}{2}}[/tex]
The integral becomes [tex]\frac{1}{a}\int u^{\frac{1}{2}}du[/tex]
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