Trigonometric substi:int (lnw)^3 / (w\sqrt{(lnw)^2+4})

[T_TEngineer_AdamT_T]

Hi
Please help me with another difficult problem;
\(\displaystyle \L \int \frac{\ln^3w}{w\sqrt{\ln^2w - 4}}dw\)

let ln(w) = 2sec(theta)
let d(ln(w)) = 2sec(theta)tan(theta)

\(\displaystyle \L\\\int \frac{(2\sec{\theta})^3dw}{e^{2\sec{\theta}}\sqrt{(2\sec{\theta})^2-4}}\)
\(\displaystyle \L\\\int \frac{8(\sec{\theta})^3(2\sec{\theta}\tan{\theta})d{\theta}}{2e^{2\sec{\theta}}\sqrt{\tan^2{\theta}}\)

cancel the tan

\(\displaystyle \L\\8\int \frac{\sec^4{\theta}d{\theta}}{e^{2\sec{\theta}}\)
^and then im stuck with this integral
should i use int by parts?

but the answer is
\(\displaystyle \L\\\frac{1}{3}\sqrt{\ln^2w-4}(8+ln^2w) + C\)

 

[pka]

Why not use parts to begin with?
\(\displaystyle \L
u = \ln ^2 (x)\;\& \;dv = \frac{{\ln (x)dx}}{{x\sqrt {\ln ^2 (x) - 4} }}\)

 

[galactus]

Hello Adam:

The integral who arrived at is not doable by elementary means, therefore, we must try something else.

First, use parts:

\(\displaystyle \L\\\int\frac{ln(w)^{3}}{w\sqrt{ln(w)^{2}-4}}dw\)

Let \(\displaystyle \L\\u=ln(w)^{2}, \;\ dv=\frac{ln(w)}{w\sqrt{ln(w)^{2}-4}}dw, \;\ v=\sqrt{ln(w)^{2}-4}, \;\ du=\frac{2ln(w)}{w}dw\)

This gives:

\(\displaystyle \L\\ln(w)^{2}\sqrt{ln(w)^{2}-4}-2\int\frac{\sqrt{ln(w)^{2}-4}ln(w)}{w}dw\)

Now, let \(\displaystyle u=ln(w), \;\ du=\frac{dw}{w}\)

\(\displaystyle \L\\\ln(w)^{2}\sqrt{ln(w)^{2}-4}-2\int{u}\sqrt{u^{2}-4}du\)

Now, let \(\displaystyle \L\\u=2sec{\theta}, \;\ du=2sec{\theta}tan{\theta}d{\theta}\)

Now continue. Okey-doke.

 

[T_TEngineer_AdamT_T]

THank u so much for the help!

 

[soroban]

Hello, T_TEngineer_AdamT_T!

An error in differentiating . . .

\(\displaystyle \L \int \frac{\ln^3w}{w\sqrt{\ln^2w - 4}}dw\)

Let: \(\displaystyle \L\,\ln w \:=\:2\cdot\sec\theta\;\;\Rightarrow\;\;w\:=\:e^{^{2\cdot\sec\theta}}\)

Then: \(\displaystyle \L\,dw\:=\:2e^{^{2\cdot\sec\theta}}\sec\theta\cdot\tan\theta\cdot d\theta\)

Substitute: \(\displaystyle \L\:\int\frac{(2\cdot\sec\theta)^3}{e^{2\cdot\sec\theta}\,\cdot\,2\cdot\tan\theta}\,\cdot\,(2e^{2\cdot\sec\theta}\sec\theta\cdot\tan\theta\cdot d\theta) \;=\;8\int\sec^4\theta\,d\theta\)

Then: \(\displaystyle \L\:8\int\sec^2\theta(\sec^2\theta\,d\theta) \;=\;8\int(\tan^2\theta\,+\,1)(\sec^2\theta\,d\theta)\)

Let \(\displaystyle u\,=\,\tan\theta\;\;\Rightarrow\;\;du\,=\,\sec^2\theta\,d\theta\)

Substitute: \(\displaystyle \L\:8\int(u^2\,+\,1)\,du \;=\;8\left(\frac{u^3}{3}\,+\,u\right)\,+\,C\;=\;\frac{8}{3}\cdot u\left(u^2\,+\,3\right)\,+\,C\)


Back-substitute: \(\displaystyle \L\:\frac{8}{3}\cdot\tan\theta(\tan^2\theta\,+\,3)\,+\,C\)


Back-substitute again
We had: \(\displaystyle \:2\cdot\sec\theta\,=\,\ln w\;\;\Rightarrow\;\;\sec\theta\,=\,\frac{\ln w}{2}\,=\,\frac{hyp}{adj}\)

. . From: \(\displaystyle \,adj\,=\,2,\:hyp\,=\,\ln w\), we get: \(\displaystyle \pp\,=\,\sqrt{\ln^2 w\,-\,4}\)

. . Hence: \(\displaystyle \L\:\tan\theta \,=\,\frac{\sqrt{\ln^2w\,-\,4}}{2}\)

And we have: \(\displaystyle \L\:\frac{8}{3}\cdot\left(\frac{\sqrt{\ln^2w\,-\,4}}{2}\right)\left[\left(\frac{\sqrt{\ln^2w\,-\,4}}{2}\right)^2\,+\,3\right]\,+\,C\)

. . \(\displaystyle \L=\;\frac{4}{3}\cdot\sqrt{\ln^2w\,-\,4}\cdot\left[\frac{\ln^2w\,-\,4}{4}\,+\,3\right]\,+\,C\)

. . \(\displaystyle \L=\;\frac{4}{3}\cdot\sqrt{\ln^2w\,-\,4}\cdot\left[\frac{\ln^2w\,+\,8}{4}\right]\,+\,C\)

. . \(\displaystyle \L=\;\frac{1}{3}\cdot\sqrt{\ln^2w\,-\,4}\cdot(\ln^2w\,+\,8)\,+\,C\)

 

[T_TEngineer_AdamT_T]

thank you so much!!