# Integral and Derivative Problems

1. Jan 14, 2009

### rwisz

1. The problem statement, all variables and given/known data
$$\int{\frac{1-sin\theta}{cos\theta}d\theta}$$

and

Find $$\frac{dy}{dx}$$ if $$y=x^ee^x$$

2. Relevant equations
NONE

3. The attempt at a solution
We've attempted every "u-substitution" we can think of for problem 1.

We've tried to differentiate problem 2 many times, and can't come up with the proper derivative.

thanks! Any advice is greatly appreciated.

2. Jan 14, 2009

### jgens

My recomendation for the first one is to split the integral into two separate functions. Then all you have to integrate is sec(x) - tan(x) which is quite simple.

For the second problem show what work you've done as it should just be an application of the product rule.

3. Jan 14, 2009

### rwisz

Thanks for the advice on Number 1! We integrated successfully!

On to number 2 however, here's what we have so far:

Using simply the product rule:
$$y'=ex^{e-1}e^x+x^ee^x$$

We've attempted to simplify down etc, but not coming up with the right answer here!
Is it perhaps using the idea a base besides e?

We're really struggling here, this is our last problem!

4. Jan 14, 2009

### NoMoreExams

Why can't you find the right answer? Factor out e^x*x^(e-1) from both parts

5. Jan 14, 2009

### rwisz

We're both looking at your reply with the most blank stare you can ever imagine...the $$x^{e-1}$$ is not in the second term whatsoever... Am I really as stupid as you're making me feel? I'm very confused here...

6. Jan 14, 2009

### jgens

As far as I can tell, the solution should be correct - aside from a restricted domain. Some thoughts on simplifications though: Factor (e^x)(x^e) out so you are left with [(e^x)(x^e)](e/x +1). That's about as simple as you can get it.

7. Jan 14, 2009

### NoMoreExams

x^e = x^(e-1) * x

8. Jan 14, 2009

### NoMoreExams

I would factor out the smallest of both powers which in this case is e-1 for the power of x to get

$$ex^{e-1}e^x+x^ee^x = e^{x}x^{e-1}\left(e + x\right)$$

9. Jan 14, 2009

### rwisz

We opted on calling our teacher as well. And she correctly guided us through the problem... and somehow it worked. I doubt she'll give us one like this on the test... but thanks for all your help, much obliged.

Thanks!

10. Jan 14, 2009

### jgens

Either works, though I suppose your approach yields a simpler expression.

11. Jan 15, 2009

### HallsofIvy

Staff Emeritus
$e^{x-1}$ certainly is in the second term: the second term includes $e^x$ and $e^x= (e^{x-1})(e)$

12. Jan 15, 2009

### NoMoreExams

You mean x^(e-1)?

13. Jan 15, 2009

### psykatic

The first integral can even be done by simple substitution i.e
$$\int{\frac{1-sin\theta}{cos\theta}d\theta}$$
can be written as,
$$\int \frac{(sin {\frac {\theta}{2}}-~cos{\frac {\theta}{2}})^2}{(cos^2{\frac {\theta}{2}}-sin^2{\frac {\theta}{2}})}$$

Last edited: Jan 15, 2009