# Integral and Derivative Problems

## Homework Statement

$$\int{\frac{1-sin\theta}{cos\theta}d\theta}$$

and

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

NONE

## 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.

jgens
Gold Member
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.

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!

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

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...

jgens
Gold Member
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.

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

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.

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

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

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!

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

HallsofIvy
Homework Helper
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...

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

You mean x^(e-1)?

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,
$$\int \frac{(sin {\frac {\theta}{2}}-~cos{\frac {\theta}{2}})^2}{(cos^2{\frac {\theta}{2}}-sin^2{\frac {\theta}{2}})}$$

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