How do you integrate #int x/sqrt(x^2-7)# by trigonometric substitution?

Question

Isaiah Allen · Accepted Answer

#int (xdx)/sqrt(x^2-7) = sqrt(x^2-7)+C#

You do not really need a trigonometric substitution here.

If you pose:

#t= x^2-7# #dt = 2xdx#

the integral becomes:

#int (xdx)/sqrt(x^2-7) = 1/2 int (dt)/sqrt(t) = sqrt(t) + C = sqrt(x^2-7)+C#

Aurora Alvarado · Answer

Although you do not need trigonometric substitution, it can be used for this integral.

Recall that #sec^2theta-1 = tan^2 theta#, so that's the basic substitution we'll use. (We could use hyperbolic functions instead, but not everyone is familiar with them when first learning trig sub.) We need, not just #sec^2 theta - 7#, but #7sec^2 theta - 7# so that we can factor out the #7# #int x/sqrt(x^2-7) dx# Let #x = sqrt7 sec theta#, so that #dx = sqrt7 sec theta tan theta d theta# and #sqrt(x^2-7) = sqrt (7sec^2 theta - 7) = sqrt7 tan theta# #int x/sqrt(x^2-7) dx = int (sqrt7sec theta)/(sqrt7 tan theta) sqrt7 sec theta tan theta d theta# # = int sqrt7 sec^2 theta d theta = sqrt7 tan theta +C# With #x = sqrt7 sec theta#, we get #tan theta = sqrt(x^2-7)/sqrt7# Therefore, #int x/sqrt(x^2-7) dx = sqrt(x^2-7) +C#