Integrate x^2/(sqrt(x^2-25)dx
let
x = 5 secθ
dx = 5 secθ tanθ
√(x^2-25) = 5 tanθ
x^2 = 25 sec^2θ
Things will get much simpler.
To integrate the given expression, x^2/(sqrt(x^2-25)), we can use a trigonometric substitution to simplify the expression.
Let's substitute x = 5sec(theta), where sec(theta) is the secant of theta.
First, let's find dx in terms of d(theta):
dx = 5sec(theta)tan(theta)d(theta)
Now, let's substitute the values of x and dx in terms of theta:
x^2 = (5sec(theta))^2 = 25sec^2(theta)
sqrt(x^2 - 25) = sqrt(25sec^2(theta) - 25) = sqrt(25(sec^2(theta) - 1)) = 5sqrt(sec^2(theta) - 1)
Now, let's rewrite the integral using the substitution:
∫ x^2/(sqrt(x^2-25))dx = ∫ (25sec^2(theta))/(5sqrt(sec^2(theta)-1))(5sec(theta)tan(theta))d(theta)
= 5 ∫ (5sec^2(theta))/(5sqrt(sec^2(theta)-1))sec(theta)tan(theta)d(theta)
Canceling out the common terms, we get:
∫ sec^2(theta)/sqrt(sec^2(theta)-1) d(theta)
Now, we can simplify the expression using trigonometric identities:
sec^2(theta) = tan^2(theta) + 1
sqrt(sec^2(theta)-1) = sqrt(tan^2(theta) + 1-1) = sqrt(tan^2(theta)) = |tan(theta)| = tan(theta)
Substituting these simplified expressions, we get:
∫ sec^2(theta)/sqrt(sec^2(theta)-1) d(theta) = ∫ tan(theta)/tan(theta) d(theta)
= ∫ 1 d(theta)
= theta + C
Finally, substituting back the original variable:
∫ x^2/(sqrt(x^2-25))dx = theta + C
Remember that we initially substituted x = 5sec(theta), so we need to express our answer in terms of x. To find theta, we can use the inverse of our substitution:
x = 5sec(theta)
sec(theta) = x/5
cos(theta) = 1/sec(theta) = 5/x
theta = arccos(5/x)
Substituting theta back into the answer, we get:
∫ x^2/(sqrt(x^2-25))dx = arccos(5/x) + C
Therefore, the integral of x^2/(sqrt(x^2-25)) with respect to x is arccos(5/x) + C, where C is the constant of integration.