Trigonometry Substitution (Integral)

[KDeep]

Homework Statement

Making sure I got the right answer.

Homework Equations

The Attempt at a Solution

 

[Dick]

Um, sec(theta)^2/sqrt(sec(theta)^2) isn't equal to 1.

 

[1MileCrash]

Hmm... I don't think so.

This doesn't need a trig sub at all. Remember your basic algebra too!

Hint: Look at the integral of the hyperbolic sine function, and see if there is something you can do to your function to make it a simple integral of that form.

 

[KDeep]

Um, sec(theta)^2/sqrt(sec(theta)^2) isn't equal to 1.

Correct, so its 1/4∫ secθ dθ

which is ln | secθ + tanθ | + C

1/4ln | (x+2/2) + (x/2) | is the answer right?

 

[Dick]

Hmm... I don't think so.

This doesn't need a trig sub at all. Remember your basic algebra too!

Hint: Look at the integral of the hyperbolic sine function, and see if there is something you can do to your function to make it a simple integral of that form.

I think it's pretty likely that KDeep hasn't done hyperbolic trig functions yet, though that does make it easier. You need some kind of substitution. Ordinary trig will work fine. You just have to do it right.

 

[KDeep]

I have never heard about hyperbolic sine function at all, I will definitely look into that. My Professor wants us to do trig substitutions since it is in the section.

 

[1MileCrash]

Correct, so its 1/4∫ secθ dθ

which is ln | secθ + tanθ | + C

1/4ln | (x+2/2) + (x/2) | is the answer right?

When you take the derivative of your solution, do you get the function you wished to integrate?

 

[Dick]

Correct, so its 1/4∫ secθ dθ

which is ln | secθ + tanθ | + C

1/4ln | (x+2/2) + (x/2) | is the answer right?

Getting close. Try the sec(arctan(x/2)) part again.

 

[KDeep]

When you take the derivative of your solution, do you get the function you wished to integrate?

I am confident that the derivative of ln| secθ + tanθ | is secθ.

 

[1MileCrash]

I am confident that the derivative of ln| secθ + tanθ | is secθ.

You are finding the antiderivative of 1/sqrt(16+4x^2), so your solution must yield 1/sqrt(16+4x^2) (or equivalent) when it is differentiated.

I think it's pretty likely that KDeep hasn't done hyperbolic trig functions yet, though that does make it easier. You need some kind of substitution. Ordinary trig will work fine. You just have to do it right.

I misspoke anyway - I meant inverse hyperbolic sine, but yes, you're right.

 

[KDeep]

You are finding the antiderivative of 1/sqrt(16+4x^2), so your solution must yield 1/sqrt(16+4x^2) (or equivalent) when it is differentiated.

I am using trigonometry substitutions though. I replaced x with 2tanθ, and dx with 2sec^2θdθ

 

[1MileCrash]

I am using trigonometry substitutions though. I replaced x with 2tanθ, and dx with 2sec^2θdθ

It doesn't matter. Given those replacements, replace them back when you have your solution.

 

[Dick]

You are almost there. You got the sec(arctan(x/2)) part wrong and you need to correct a simple numerical mistake. That's all.