Posted by **Jason** on Thursday, September 22, 2011 at 2:23pm.

Can someone help me to evaluate these two integrals?

INT dx/(x^2*sqrt(x^2-25))

and

INT (from 0 to 1) -5*sqrt(x^2+1) dx

- Calculus -
**Steve**, Thursday, September 22, 2011 at 3:20pm
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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