# Trigonometry Substitution (Integral)

1. Feb 22, 2013

### KDeep

1. The problem statement, all variables and given/known data
Making sure I got the right answer.

2. Relevant equations

3. The attempt at a solution

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2. Feb 22, 2013

### Dick

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

3. Feb 22, 2013

### 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.

4. Feb 22, 2013

### KDeep

Correct, so its 1/4∫ secθ dθ

which is ln | secθ + tanθ | + C

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

5. Feb 22, 2013

### Dick

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.

6. Feb 22, 2013

### 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.

7. Feb 22, 2013

### 1MileCrash

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

8. Feb 22, 2013

### Dick

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

9. Feb 22, 2013

### KDeep

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

10. Feb 22, 2013

### 1MileCrash

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

11. Feb 22, 2013

### KDeep

12. Feb 22, 2013

### 1MileCrash

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

13. Feb 22, 2013

### 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.