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Edit: The integral needs to be solved with the exact value of u given below.

I am asked to solve the following indefinite integral:

$\int\frac{e^{6x}}{\sqrt{9-e^{12x}}}dx$

By making the substitution $u=\frac{e^{6x}}3$

The first thing I need to do is get the integrand in terms of u, but I'm having trouble actually performing the substitution. I'm guessing the 3 in the u-sub will affect the square root but I can't quite grasp how.

So far I've just calculated the following:

$\frac{du}{dx}=2e^{6x}$ , $\ du = 2e^{6x}dx$ , $\ dx = \frac{du}{2e^{6x}}$

Then I just subbed in the value of dx to the original integral to get:

$\int\frac{e^{6x}}{2e^{6x}\sqrt{9-e^{12x}}}du$

From here I still cannot see how I am supposed to perform the substitution, or if I was better off subbing in u before the value of dx.

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    $\begingroup$ But you have $e^{6x}=3u$ and hence $e^{12x}=9u^2$. $\endgroup$ – almagest May 22 '16 at 12:18
  • $\begingroup$ Alright that cleared things up a lot. You can put it as an answer if you want. $\endgroup$ – Dwayne H May 22 '16 at 12:30
  • $\begingroup$ Someone has already done that. Are you happy where to go from there? $\endgroup$ – almagest May 22 '16 at 12:32
  • $\begingroup$ Yep. Whole thing is done now. Was just the initial mental blank that threw me. Cheers. $\endgroup$ – Dwayne H May 22 '16 at 12:33
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$$\int\frac{e^{6x}}{\sqrt{9-e^{12x}}}\space\text{d}x=$$


Substitute $u=e^x$ and $\text{d}u=e^x\space\text{d}x$:


$$\int\frac{u^5}{\sqrt{9-u^{12}}}\space\text{d}u=$$


Substitute $s=u^6$ and $\text{d}s=6u^5\space\text{d}u$:


$$\frac{1}{6}\int\frac{1}{\sqrt{9-s^2}}\space\text{d}s=\frac{1}{6}\int\frac{1}{3\sqrt{1-\frac{s^2}{9}}}\space\text{d}s=$$


Substitute $p=\frac{s}{3}$ and $\text{d}p=\frac{1}{3}\space\text{d}s$:


$$\frac{1}{6}\int\frac{1}{\sqrt{1-p^2}}\space\text{d}p=\frac{\arcsin\left(p\right)}{6}+\text{C}=\frac{\arcsin\left(\frac{e^{6x}}{3}\right)}{6}+\text{C}$$

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  • $\begingroup$ On second thought, this answer isn't using the u value that I put in the question is it? $\endgroup$ – Dwayne H May 22 '16 at 12:37
  • $\begingroup$ Indeed it isn't! $\endgroup$ – Jan May 22 '16 at 12:43
  • $\begingroup$ I thought so. The marking system wont accept anything as correct if it isn't using the same value for u, unfortunately. Otherwise I could've done this much easier. $\endgroup$ – Dwayne H May 22 '16 at 12:45
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Observe that by the chain rule:

$$\int\frac{dx}{\sqrt{1-x^2}}=\arcsin x+C\implies \int\frac{f'(x)\,dx}{\sqrt{1-f(x)^2}}=\arcsin f(x)+C$$

Now:

$$9-e^{12x}=9\left(1-\left(\frac{e^{6x}}3\right)^2\right)\;,\;\;\text{and}\;\;\left(\frac{e^{6x}}3\right)'=2e^{6x}\implies$$

$$\int\frac{e^{6x}}{\sqrt{9-e^{12x}}}dx=\frac16\int\frac{2e^{6x}dx}{\sqrt{1-\left(\frac{e^{6x}}3\right)^2}}=\frac16\arcsin\frac{e^{6x}}3+C$$

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