Integration By Trig Substitution: 1/sqrt(x^2 -4)?

Question

I'm using trig substitution to solve this problem. I know that x=2sec(theta) and so I know from this that dx=2sec(theta)tan(theta). Through trig I get that tan(theta)=sqrt(x^2 -4)/2. Using substitution I get the integral to equal ln|sec(theta) + tan(theta)| and resubstituting I get ln|x/2 + sqrt(x^2 -4)/2|. However the answer is ln|x + sqrt(x^2 -4)|. Where did I go wrong?

Answer 1

I'll solve it too and you can compare steps with mine.

Integral (1/sqrt(x^2 - 4))dx

Let x = 2sec(t)
dx = 2sec(t)tan(t)

Integral ( [1/sqrt(4sec^2(t) - 4)] [2sec(t)tan(t)] ) dt

(skipping details)

Integral ( [2sec(t)tan(t)] / [2tan(t)] )dt

Integral ( sec(t) ) dt

= ln |sec(t) + tan(t)| + C

Since we let x = 2sec(t), it follows that sec(t) = x/2.
By SOHCAHTOA, since cos(t) = adj/hyp, sec(t) = hyp/adj.

Make a right angle triangle with angle t, and hypotenuse equal to x, and adjacent equal to 2. It follows that opposite is equal to sqrt(x^2 - 4).
hyp = x, adj = 2, opp = sqrt(x^2 - 4)

It follows that tan(t) = opp/adj = sqrt(x^2 - 4)/2, so

ln |sec(t) + tan(t)| + C = ln|(x/2) + sqrt(x^2 - 4)/2| + C

We got the same answer. I can tell you why they have a different answer (your answer is not wrong).

Note that we can factor out (1/2) inside the ln.

ln| (1/2) (x + sqrt(x^2 - 4)) | + C

By the log property log[base b](ac) = log[base b](a) + log[base b](c), we have

ln(1/2) + ln|x + sqrt(x^2 - 4)| + C

Rearranging this,

ln|x + sqrt(x^2 - 4)| + ln(1/2) + C

Note that ln(1/2) is a constant, and can be merged with the existing constant (since a constant plus a constant is just a constant. So we might as well just eliminate that ln(1/2), and our answer is

ln|x + sqrt(x^2 - 4)| + C

If we REALLY wanted to be picky, we can even express C as
ln(e^c), and have

ln|x + sqrt(x^2 - 4)| + ln(e^c)

And we can use the log property to merge these two into

ln | (e^c) (x + sqrt(x^2 - 4)) |

But e^c is just some constant K, so our answer is

ln | K (x + sqrt(x^2 - 4)) |

Logs do funny things with constants. In any case, you did nothing wrong, and should not lose any marks if doing this on a final exam.

Answer 2

Just a quick thought: the indefinite integral is only determined modulo an additive constant. Since your answer differs from the one you say is right by just a factor of 1/2 inside the square root, you can remove it:
ln(x/2 + sqrt(x^2 -4)/2) = ln(1/2*(x + sqrt(x^2 -4))) = ln(1/2) + ln(x + sqrt(x^2 -4)), so your answer differs only by an additive constant, therefor it is the same answer.

Answer 3

cos(x) * dx / (a million + sin(x)^2)^(3/2) t = sin(x) dt = cos(x) * dx dt / (a million + t^2)^(3/2) t = tan(m) dt = sec(m)^2 * dm sec(m)^2 * dm / (a million + tan(m)^2)^(3/2) => sec(m)^2 * dm / (sec(m)^2)^(3/2) => sec(m)^2 * dm / sec(m)^3 => dm / sec(m) => cos(m) * dm combine sin(m) + C returned-exchange t = tan(m) t = sin(m) / cos(m) t * cos(m) = sin(m) t^2 * cos(m)^2 = sin(m)^2 t^2 * (a million - sin(m)^2) = sin(m)^2 t^2 - t^2 * sin(m)^2 = sin(m)^2 t^2 = sin(m)^2 * (a million + t^2) sin(m)^2 = t^2 / (a million + t^2) sin(m) = t / sqrt(a million + t^2) sin(m) + C => t / sqrt(a million + t^2) + C t = sin(x) sin(x) / sqrt(a million + sin(x)^2) + C