# Thread: help with integrating fractions with e in them

1. ## help with integrating fractions with e in them

Hey everyone, first post ever in here. This is my 2nd time taking Calculus 2 and I feel like I am worse off than the first time.

I have a quick set of 4 problems that Im not sure I understand how to do. Apparently 1, 2 and 3 can be solved with u-substitution and a formula, while 4 just needs a u-substitution.

Here are the problems, sorry about the format:

1: $\int\frac{1}{\sqrt{9-e^{2x}}}\,dx$

2: $\int\frac{e^x}{\sqrt{9-e^{2x}}}\,dx$

3: $\int\frac{1}{e^x\sqrt{9-e^{2x}}}\,dx$

4: $\int\frac{e^2x}{\sqrt{9-e^{2x}}}\,dx$

I feel like Ive lost a lot of my fundemental calc skills.

-WhiteX

2. Hello, whitex!

Welcome aboard!
I'll run through the first one . . . it takes quite a bit of work.

$1)\;\;\int \frac{dx}{\sqrt{9-e^{2x}}}$
We use a Trig Substitution . . .

Let: $e^x \:=\:3\sin\theta \quad\Rightarrow\quad x \:=\:\ln(3\sin\theta) \quad\Rightarrow\quad dx \:=\:\frac{3\cos\theta}{3\sin\theta}d\theta \:=\:\cot\theta\,d\theta$

. . and: . $\sqrt{9-e^{2x}} \:=\:\sqrt{9-9\sin^2\!\theta} \:=\:\sqrt{9(1-\sin^2\!\theta)} \:=\:\sqrt{9\cos^2\!\theta} \:=\:3\cos\theta$

Substitute: . $\int\frac{\cot\theta\,d\theta}{3\cos\theta} \;=\;\frac{1}{3}\int\frac{\frac{\cos\theta}{\sin\t heta}}{\cos\theta}\,d\theta \;=\;\frac{1}{3}\int\frac{d\theta}{\sin\theta} \;=\;\frac{1}{3}\int\csc\theta\,d\theta$

Integrate: . $\frac{1}{3}\ln|\csc\theta - \cot\theta| + C$ .[1]

Back-substitute: . $3\sin\theta \:=\:e^x\quad\Rightarrow\quad \sin\theta \:=\:\frac{e^x}{3} \:=\:\frac{opp}{hyp}$

We see that $\theta$ is in a right triangle with: . $opp = e^x,\;hyp = 3$

. . Using Pythagorus, we have: . $adj = \sqrt{9-e^{2x}}$

Hence: . $\csc\theta \:=\:\frac{3}{e^x},\;\cot\theta \:=\:\frac{\sqrt{9-e^{2x}}}{e^x}$

Substitute into [1]: . $\frac{1}{3}\ln\left|\frac{3}{x} - \frac{\sqrt{9-e^{2x}}}{e^x}\right| + C \;\;=\;\;\frac{1}{3}\ln\left|\frac{3-\sqrt{9-e^{2x}}}{e^x}\right| + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This answer can be simplified further . . .

. . $\frac{1}{3}\bigg[\ln\left|3 - \sqrt{9-e^{2x}}\right| - \ln\left(e^x\right)\bigg] + C \;\;=\; \;\frac{1}{3}\bigg[\ln\left|3-\sqrt{9-e^{2x}}\right| - x\bigg] + C$

3. Wow thanks I appreciate it.

This site is great. Any way you want to assist me in the other 3? If not, thats ok!

4. Hello, whitex!

#3 and #4 are similar to the first one.
#2 is the easiest of them.

$2)\;\;\int\frac{e^x\,dx}{\sqrt{9-e^{2x}}}$

Let: $u \:=\:e^x\quad\Rightarrow\quad du \:=\:e^x\,dx$

Substitute: . $\int\frac{ du}{\sqrt{9-u^2}} \;=\;\sin^{-1}\left(\tfrac{u}{3}\right) + C$

Back-substitute: . $\sin^{-1}\left(\frac{e^x}{3}\right) + C$

$3)\;\;\int\frac{dx}{e^x\sqrt{9-e^{2x}}}$
This one uses the same substitution as #1 . . .

Let: $e^x \:=\:3\sin\theta \quad\Rightarrow\quad x \:=\:\ln(3\sin\theta) \quad\Rightarrow\quad dx \:=\:\cot\theta\,d\theta$
. . and: . $\sqrt{9-e^{2x}} \:=\:3\cos\theta$

Substitute: . $\int\frac{\cot\theta\,d\theta}{3\sin\theta\cdot3\c os\theta} \;=\;\frac{1}{9}\int\frac{d\theta}{\sin^2\!\theta} \;=\; \frac{1}{9}\int\csc^2\!\theta\,d\theta \;=\; -\frac{1}{9}\cot\theta + C$

Back-substitute: . $e^x \:=\:3\sin\theta \quad\Rightarrow\quad\sin\theta \:=\:\frac{e^x}{3} \:=\:\frac{opp}{hyp}$
. . Then: . $adj \:=\:\sqrt{9-e^{2x}}$

And we have: . $-\frac{1}{9}\,\frac{\sqrt{9-e^{2x}}}{e^x} + C$

$4)\;\;\int\frac{e^{2x}\,dx}{\sqrt{9-e^{2x}}}$
Same subsitution as #3 . . .

Substitute: . $\int\frac{(9\sin^2\!\theta)(\cot\theta\,d\theta)}{ 3\cos\theta} \;=\;3\int\sin\theta\,d\theta \;=\;-3\cos\theta + C$

Back-substitute: . $-\frac{1}{3}\cdot\frac{\sqrt{9-e^{2x}}}{3} + C \;=\;-\sqrt{9-e^{2x}} + C$