# algebraic/trigonometric substitution

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• December 6th 2007, 08:33 AM
cazimi
algebraic/trigonometric substitution
$\int\frac{e^-x}{(9e^-2x+1)^3/2}\, dx$

$\int\frac{dz}{(z^2-6z+18)^3/2}$

$\int\frac{ln^3w}{w\sqrt{ln^2w-4}}\,dw$
• December 6th 2007, 09:11 AM
Krizalid
Quote:

Originally Posted by cazimi
$\int\frac{e^{-x}}{(9e^{-2x}+1)^{3/2}}\, dx$

Substitute $u=e^{-x},$

$\int {\frac{{e^{ - x} }}
{{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = - \int {\frac{{du}}
{{\left( {9u^2 + 1} \right)\sqrt {9u^2 + 1} }}} .$

Make another substitution defined by $u=\frac1\tau,$

$\int {\frac{{e^{ - x} }}
{{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = \int {\frac{\tau }
{{\left( {9 + \tau ^2 } \right)\sqrt {9 + \tau ^2 } }}\,d\tau } .$

One more substitution according to $\varphi = \sqrt {9 + \tau ^2 } ,$

$\int {\frac{{e^{ - x} }}
{{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = \int {\varphi ^{ - 2} \,d\varphi } = - \frac{1}
{\varphi } + k.$

Back substitute:

\begin{aligned}
\varphi &= \sqrt {9 + \frac{1}
{{u^2 }}}\\
&= \sqrt {9 + \frac{1}
{{e^{ - 2x} }}}\\
&= \frac{{\sqrt {9e^{ - 2x} + 1} }}
{{e^{ - x} }}.
\end{aligned}

And we happily get $\int {\frac{{e^{ - x} }}
{{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = - \frac{{e^{ - x} }}
{{\sqrt {9e^{ - 2x} + 1} }} + k.$
• December 6th 2007, 10:18 AM
TwistedOne151
Second one
On the second integral,

First, I think you meant
$\int\frac{dz}{(z^2-6z+18)^{3/2}}$, not $\int\frac{dz}{(z^2-6z+18)^3/2}$ (you forgot to bracket the exponent).

First, we want to complete the square in the denominator: $z^2-6z+18=z^2-6z+9+9=(z-3)^2+9$

So using $u=z-3$, we get $\int\frac{dz}{(z^2-6z+18)^{3/2}} = \int\frac{du}{(u^2+3^2)^{3/2}}$

From here, you can use the trigonometric substitution $u=3 \tan \theta$, which gives $u^2+3^2 = 3^2 (\tan^2 \theta +1) = (3 \sec \theta)^2$ and $du = 3 \sec^2 \theta d\theta$
And thus
$\int\frac{dz}{(z^2-6z+18)^{3/2}} = \int\frac{3 \sec^2 \theta d\theta}{(3 \sec \theta)^3} = \int\frac{d\theta}{9 \sec \theta} = \frac{1}{9} \int \cos \theta d\theta$

You can take it from there.

--Kevin C.
• December 6th 2007, 10:25 AM
PaulRS
Well, the third

Let $u=\ln(w)$ thus: $\frac{du}{dw}=\frac{1}{w}$

Then $\int\frac{ln^3w}{w\sqrt{ln^2w-4}}dw=\int\frac{u^3}{\sqrt{u^2-4}}du$

Now $z=\sqrt{u^2-4}$ so: $\frac{dz}{du}=\frac{u}{\sqrt{u^2-4}}$

Therefore: $\int\frac{u^3}{\sqrt{u^2-4}}du=\int(z^2+4)dz$