Posted by Jason on Thursday, September 22, 2011 at 2:23pm.
Run, do not walk to the section on trig substitutions. Usually that's the best tool for problems like these.
Recall that sec^2 = 1 + tan^2, so that
sec^2 - 1 = tan^2. Let x=5sec(u), so x^2-25 = 25sec^2(u)-25 = 25 tan^2(u)
x^2 = 25 sec^2(u)
dx = 5 sec(u) tan(u)
Our integral now becomes
5sec(u)tan(u)/(25sec^2(u) * 5tan(u)) = 1/25sec(u) du = cos(u)/25 du
Int 1/25 cos(u) = 1/25 sin(u)
Since x = 5 sec(u),
x/5 = sec(u), and 5/x = cos(u)
sin(u) = sqrt(1-25/x^2) = 1/x * sqrt(x^2-25), and our answer is thus
1/25 sin(u) = sqrt(x^2-25)/25x
*whew*
The second one is similar, but uses hyperbolic functions:
Let x = tan(u), so x^2+1 = sec^2(u)
dx = sec^2(u) du
and the integral becomes
-5 Int sec^3(u) du
Use integration by parts (twice) to get
-5/2 tan(u)sec(u) - 5/2 log(tan u + sec u)
= -5/2 x sqrt(x^2+1) - 5/2 log(x + sqrt(x^2+1)
= -5/2[x sqrt(x^2+1) + arcsinh(x)]
from 0 to 1
At 0, the function is 0, so we just have f(1) = -5/2(sqrt(2) + asinh(1))
= -5/2(1.414 + 2.435) = 4.62
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