The integral is $$\int \frac{1}{\sqrt{2x}}\ dx$$
The assignment says use $u$-substitution, but I don't know what I should define $u$ as. Any help would be appreciated, thanks!
Ohh, I get it now! Thanks to everyone who answered!
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3 Answers
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Start by factoring out the $\sqrt{2}$, such that your integral is: $$I=\dfrac{1}{\sqrt{2}}\int\dfrac{1}{\sqrt{x}}dx$$ Now make the substitution: $$u=\sqrt{x}$$ $$du=\dfrac{dx}{2\sqrt{x}}$$ Such that: $$I=\dfrac{2}{\sqrt{2}}\int du=\sqrt{2}\ u$$ And finally, substituting $x$ back: $$\int\dfrac{1}{\sqrt{2x}}dx=\sqrt{2x}$$
answered Mar 12, 2015 at 21:59
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$\begingroup$ ..why did you bother doing a substitution??? $\endgroup$ Mar 12, 2015 at 22:39
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$\begingroup$ @JpMcCarthy Because we're obviously not assuming that the integral of $x^{-1/2}$ is a known result. However, it is fair to say that the derivative of $\sqrt{x}$ should be. Substitution moves the problem from an integral to a derivative. $\endgroup$ Mar 13, 2015 at 9:29
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$\begingroup$ I would imagine that antidifferentiation of $x^r$ is always presented before $u$-substitution....I mean we differentiate $x^r$ using implicit differentiation...we know what to do with $\int x^r\,dx$...for rational $r$. $\endgroup$ Mar 13, 2015 at 9:58
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Set $u=2x$.Then the integral is transformed to $\frac {1}{2} \int \frac {du}{\sqrt{u}}$ which is now easy to solve.
answered Mar 12, 2015 at 21:56
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I would use $2x = u^2$ Then you get $$ \int \frac{1}{\sqrt{u^2}}udu = \int du = u + C $$ Thus $$ \sqrt{2x} + C $$
