Integration with trig substitution

[Ethan Godden]

Homework Statement

The problem is the integral attached

Homework Equations

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

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.

Thank you in advance,

Ethan

 

[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.

 

[SammyS]

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.

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

 

[Ethan Godden]

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

 

[Mark44]

Homework Statement

The problem is the integral attached

Homework Equations

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

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.

Thank you in advance,

Ethan

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.

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.

No, that would be plain old sec(x).

 

[Mark44]

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

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

 

[Ethan Godden]

Thank you,

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.

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

 

[Mark44]

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.

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).

 

[Ethan Godden]

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).

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)?

 

[Mark44]

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)?

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##.

 

[vela]

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.