# Homework Help: Integration with trig substitution

1. Mar 7, 2016

### Ethan Godden

1. The problem statement, all variables and given/known data
The problem is the integral attached

2. Relevant equations
sec2(u)=(1+tan2(x))
a2+b2=c2
∫cos(u)=-sin(u)+C

3. The attempt at a solution
The solution is attached. I am wondering if someone could give me a hint where I went drastically wrong or where I possibly dropped a negative. Hopefully what will come of this is the latter.

Ethan

File size:
489.4 KB
Views:
53
2. Mar 7, 2016

### Ethan Godden

The one idea I have, by the way, is when I simplify the √sec2(x), it could become -sec(x). This is my only idea.

3. Mar 7, 2016

### SammyS

Staff Emeritus
Yes, if sec(x) < 0, then $\ \sqrt{\sec^2(x)\,}=-\sec (x)\ .$

4. Mar 7, 2016

### Ethan Godden

how do I know sec(x)< 0? The only information given is the integral at the beginning and that a>0.

5. Mar 7, 2016

### Staff: Mentor

I see two mistakes.
1. $\int cos u du = sin u$, not -sin u
2. $sec u = \frac {\sqrt{a^2 + x^2}}{a}$. You don't have that written down, and I suspect that you lost a factor of a in your simplification.
No, that would be plain old sec(x).

6. Mar 7, 2016

### Staff: Mentor

If $0 \le u < \pi/2$, sec(u) will be positive. This is a reasonable assumption in your trig substitution.

7. Mar 7, 2016

### Ethan Godden

Thank you,

My only follow up question is why is this significant?

I am not at sin(u)/a. I know by the trig circle that the opposite side is equal to x, the adjacent side is a, and the hypotenuse is √(x2+a2 ). This means sin(u)=x/√(x2+a2 ). I don't know where the sec(u) back substitution came from.

Thank You,

Ethan

8. Mar 8, 2016

### Staff: Mentor

I always draw a triangle when I do trig substitution -- I don't clutter up my brain with memorized forumulas that I might forget. In my triangle u is the acute angle I'm interested in. The adjacent side is a, the opposite side is x, the hypotenuse is $\sqrt{a^2 + x^2}$. From this triangle I get $tan u = \frac x a$ and my expression for sec(u).

9. Mar 8, 2016

### Ethan Godden

I guess my main question is why are you dealing with sec(u) when the question ends up with sin(u)/a2. Shouldn't I be drawing a trig ciricle for sin(u) and not sec(u)?

10. Mar 8, 2016

### Staff: Mentor

The sec(u) factor comes in because of the $(a^2 + x^2)^{3/2}$ in the denominator. In the triangle I described, $\tan u = \frac x a$ and $\sec u = \frac{\sqrt{a^2 + x^2}}{a}$. I hope that's clear.

You use the triangle (not a trig "ciricle") to undo the substitution $\tan u = \frac x a$. From the triangle you can figure out all of the trig relationships, including $\sin u$.

11. Mar 8, 2016

### vela

Staff Emeritus
I would suggest omitting some of the unnecessary algebraic manipulations, like what you did with the denominator before doing the trig substitution.
$$(a^2+x^2)^{3/2} = (\sqrt{a^2+x^2})^3 = (\sqrt{a^2+a^2\tan^2 u})^3 = (a\sec u)^3$$ Fewer steps, easier to read for the grader, less chance of making a dumb algebra mistake, etc. Also, after you change variables to $u$, the limits are no longer $0$ and $a$. If you had changed the limits to $u=0$ to $u=\pi/4$, you probably would have realized that $\sec u > 0$ and that your error lay elsewhere.