How do you integrate #int sqrt(7+5x^2)/x# using trig substitutions?

1 Answer
Jan 10, 2017

#I=intsqrt(7+5x^2)/xdx#

We will use the substitution #sqrt5x=sqrt7tantheta#. Differentiating this shows that #sqrt5dx=sqrt7sec^2thetad theta#. Then:

#I=1/sqrt5intsqrt(7+(sqrt5x)^2)/xsqrt5dx#

Substituting in what we know (recall that #x=sqrt7/sqrt5tantheta#):

#I=1/sqrt5intsqrt(7+(sqrt7tantheta)^2)/(sqrt7/sqrt5tantheta)sqrt7sec^2thetad theta#

Clearing up the square root, and moving around the constants:

#I=1/sqrt5sqrt5/sqrt7sqrt7intsqrt(7+7tan^2theta)/tanthetasec^2thetad theta#

All the constants outside the integral cancel. We can factor #sqrt7# from the square root:

#I=int(sqrt7sqrt(1+tan^2theta))/tanthetasec^2thetad theta#

Since #1+tan^2theta=sec^2theta#:

#I=sqrt7intsqrt(sec^2theta)/tanthetasec^2thetad theta=sqrt7intsec^3theta/tanthetad theta#

Writing as #sintheta# and #costheta#:

#I=sqrt7int1/cos^3thetacostheta/sinthetad theta=sqrt7int1/cos^2theta1/sinthetad theta=sqrt7intsec^2theta/sinthetad theta#

Rewriting (again)!

#I=sqrt7int(tan^2theta+1)/sinthetad theta=sqrt7int(tan^2theta/sintheta+1/sintheta)d theta#

Rearranging...

#I=sqrt7int(sin^2theta/cos^2theta1/sintheta+csctheta)d theta#

#I=sqrt7intsintheta/costheta1/costhetad theta+sqrt7intcscthetad theta#

#I=sqrt7inttanthetasecthetad theta+sqrt7intcscthetad theta#

These are common integrals:

#I=sqrt7sectheta-sqrt7lnabs(csctheta+cottheta)+C#

We started with the substitution #sqrt5x=sqrt7tantheta#. This implies that #tantheta=(sqrt5x)/sqrt7#. In a right triangle, this means we have an opposite side of #sqrt5x# and an adjacent side of #sqrt7#. Through the Pythagorean theorem, we see that the hypotenuse is #sqrt(7+5x^2)#.

Using our definitions of the trigonometric functions, we see that:

  • #sectheta="hypotenuse"/"adjacent"=sqrt(7+5x^2)/sqrt7#
  • #csctheta="hypotenuse"/"opposite"=sqrt(7+5x^2)/(sqrt5x)#
  • #cottheta="adjacent"/"opposite"=sqrt7/(sqrt5x)#

Then:

#I=sqrt7sqrt(7+5x^2)/sqrt7-sqrt7lnabs(sqrt(7+5x^2)/(sqrt5x)+sqrt7/(sqrt5x))+C#

#I=sqrt(7+5x^2)-sqrt7lnabs((sqrt(7+5x^2)+sqrt7)/(sqrt5x))+C#

Note that #1/sqrt5# can be pulled out of the logarithm through the rule #log(a/b)=log(a)-log(b)#. It will be absorbed into the constant of integration, so this is most simply written:

#I=sqrt(7+5x^2)-sqrt7lnabs((sqrt(7+5x^2)+sqrt7)/x)+C#