Indefinite trigonometric integral with an Nth Root

[WalrusMunchies]

Homework Statement

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

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 into make "a" a linear constant.

 

[Dick]

Homework Statement

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

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}\,dx = {-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 into make "a" a linear constant.

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.

 

[WalrusMunchies]

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.

Thanks for the reply

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.

Normally I would see my professor about this, but it is reading week.

 

[Dick]

Thanks for the reply

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.

Normally I would see my professor about this, but it is reading week.

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.

 

[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)

 

[SammyS]

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

Maybe the website would like this better.

 

[Dick]

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

Maybe the website would like this better.

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