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May 19th, 2017, 11:06 AM   #1
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Stuck on this integral

Hi guys.

What is the best way to integrate this?

I can't seem to wrap my head around it.
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Last edited by skipjack; May 19th, 2017 at 11:48 PM.
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May 19th, 2017, 11:34 AM   #2
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if the integrand is $\left(x + \dfrac{5}{9}\right)^{3/2}$, then the antiderivative is

$\dfrac{2}{5}\left(x + \dfrac{5}{9}\right)^{5/2} + C$

if the integrand is $\left(\dfrac{x + 5}{9}\right)^{3/2} = \dfrac{(x+5)^{3/2}}{27}$, then the antiderivative is

$\dfrac{2}{135}\left(x + 5 \right)^{5/2} + C$
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May 19th, 2017, 11:53 AM   #3
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It looks ideal for a simple $u$ substitution.
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May 20th, 2017, 02:41 AM   #4
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To try to find $\displaystyle \!\int\! f(x)dx$ using the substitution $x = g(u)$, where $g(u)$ is invertible and differentiable,
find $\dfrac{dx}{du} = g^\prime\!(u)$, then use $\displaystyle \!\int\! f(x)dx = \!\int\! f(g(u))\frac{dx}{du}du = \!\int\! f(g(u))g^\prime\!(u)du$.

After integrating, write the result in terms of $x$ (using the inverse of $g$ if necessary).

Explicitly specifying the domain of $g$ is advisable to assist with that and simplifying $f(g(u))g^\prime\!(u)$.

To find, for example, $\displaystyle \!\int\! \frac{1}{x^2\sqrt{1 - x^2}} dx$, where $0 < x < 1$,
let $x = \cos(u)$, where $0 < u < \pi/2$, so that $\dfrac{dx}{du} = -\sin(u)$ and $\sqrt{1 - x^2} = \sin(u)$.

This gives $\displaystyle \!\int\! \frac{1}{\cos^2(u)\sin(u)} (-\sin(u)) du = \!\int\! -\sec^2(u)du = -\tan(u) + \text{C} = -\frac{\sin(u)}{\cos(u)} + \text{C}$ (where $\text{C}$ is a constant),
which equals $-\dfrac{\sqrt{1 - x^2}}{x} + \text{C}$.
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