# Indefinite trigonometric integral with an Nth Root

1. Feb 16, 2014

### WalrusMunchies

1. The problem statement, all variables and given/known data

Solve: $\int sin(16x) \sqrt[a]{cos(16x)}\,dx$ Answer should be linear in the constant "a"

3. The attempt at a solution

$\int sin(16x) \sqrt[a]{cos(16x)}\,dx$ Set: $u=cos(16x), du=-16sin(16x) du$ $~~\Rightarrow~~ {-1/16}\int \sqrt[a]{u}\,du = {-1/16}(\frac{1}{1+1/a}u^{1+{1/a}})+C = {-1/16}(\frac{1}{1+1/a}cos^{1+{1/a}}(16x))+C$

This is what I did initially, then i realized it wasn't linear in the constant "a".
I have a feeling i may need to substitute a log function in to make "a" a linear constant.

Last edited: Feb 16, 2014
2. Feb 16, 2014

### Dick

What you did is fine, except you mixed up some dx's and du's in the presentation. I have no idea what they mean by "Answer should be linear in the constant a". It's clearly not.

3. Feb 16, 2014

### WalrusMunchies

The question initially doesn't say "Answer should be linear in the constant a", but my homework website doesn't accept "+C" in this question, and gives me that message as a response when i put in my answer.

4. Feb 16, 2014

### Dick

Well, there should be a "+C" in the answer if it's an indefinite integral, and I think you did it correctly. Not sure what to say.

5. Feb 16, 2014

### scurty

I can only think of two things. The site might require a more standard form after simplifying, and it's possible you inputted the answer incorrectly. How do you input your answers? Is it text based or is there a GUI for you to use?

For example, if you input cos^(1/2)(x) into WolframAlpha, it reads it as $\sqrt{\cos{x}} \cdot x$ where you might have wanted it to be intepretted as simply $\sqrt{\cos{x}}$.

(I brought up this example because this is how you formatted your answer to the problem)

6. Feb 16, 2014

### SammyS

Staff Emeritus
$\displaystyle \frac{1}{1+1/a} = \frac{a}{a+1}$

Maybe the website would like this better.

7. Feb 16, 2014

### Dick

Well, that's not "linear in a" either. Who knows what the website would like?