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

2 Answers
Jan 11, 2017

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

Explanation:

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#

Jan 11, 2017

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

Explanation:

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#