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

Question

William Abdalla · Accepted Answer

#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:

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#

Charles Anctil · Answer

undefined To integrate $ \frac{\sqrt{7 + 5x^2}}{x} $ using trigonometric substitution, we can let $ x = \sqrt{\frac{7}{5}} \tan(\theta) $. Then, $ dx = \sqrt{\frac{7}{5}} \sec^2(\theta) d\theta $. Substituting these into the integral, we get:

$$ \int \frac{\sqrt{7 + 5x^2}}{x} dx = \int \frac{\sqrt{7 + 5\left(\sqrt{\frac{7}{5}} \tan(\theta)\right)^2}}{\sqrt{\frac{7}{5}} \tan(\theta)} \sqrt{\frac{7}{5}} \sec^2(\theta) d\theta $$

$$ = \int \frac{\sqrt{7 + 5\frac{7}{5}\tan^2(\theta)}}{\tan(\theta)} \sqrt{\frac{7}{5}} \sec^2(\theta) d\theta $$

$$ = \int \frac{\sqrt{7(1 + \tan^2(\theta))}}{\tan(\theta)} \sqrt{\frac{7}{5}} \sec^2(\theta) d\theta $$

$$ = \int \frac{\sqrt{7 \sec^2(\theta)}}{\tan(\theta)} \sqrt{\frac{7}{5}} \sec^2(\theta) d\theta $$

$$ = \int \frac{\sqrt{7} \sec(\theta)}{\tan(\theta)} \sqrt{\frac{7}{5}} \sec^2(\theta) d\theta $$

$$ = \int \sqrt{\frac{7}{5}} \frac{\sqrt{7} \sec(\theta)}{\sin(\theta)} \sec^2(\theta) d\theta $$

$$ = \int \sqrt{\frac{7}{5}} \frac{\sqrt{7} \sec^3(\theta)}{\sin(\theta)} d\theta $$

$$ = \int \sqrt{\frac{7}{5}} \frac{\sqrt{7} \sec(\theta) \sec^2(\theta)}{\sin(\theta)} d\theta $$

$$ = \int \sqrt{\frac{7}{5}} \frac{\sqrt{7} \sec(\theta) \sec(\theta) \tan(\theta)}{\sin(\theta)} d\theta $$

$$ = \int \sqrt{\frac{7}{5}} \frac{7 \tan(\theta)}{\sin(\theta)} d\theta $$

$$ = \int \sqrt{\frac{7}{5}} \frac{7}{\cos(\theta)} d\theta $$

$$ = \int \frac{7\sqrt{7}}{\sqrt{5} \cos(\theta)} d\theta $$

$$ = \frac{7\sqrt{7}}{\sqrt{5}} \int \frac{1}{\cos(\theta)} d\theta $$

$$ = \frac{7\sqrt{7}}{\sqrt{5}} \int \sec(\theta) d\theta $$

$$ = \frac{7\sqrt{7}}{\sqrt{5}} \ln|\sec(\theta) + \tan(\theta)| + C $$

$$ = \frac{7\sqrt{7}}{\sqrt{5}} \ln|\sqrt{\frac{7}{5}} \tan(\theta) + \tan(\theta)| + C $$

$$ = \frac{7\sqrt{7}}{\sqrt{5}} \ln|\tan(\theta)\left(1 + \sqrt{\frac{7}{5}}\right)| + C $$

$$ = \frac{7\sqrt{7}}{\sqrt{5}} \ln\left|\tan(\theta) + \sqrt{\frac{7}{5}}\tan(\theta)\right| + C $$

$$ = \frac{7\sqrt{7}}{\sqrt{5}} \ln|\sqrt{\frac{7}{5}}x + x| + C $$

$$ = \frac{7\sqrt{7}}{\sqrt{5}} \ln|x(\sqrt{\frac{7}{5}} + 1)| + C $$

$$ = \frac{7\sqrt{7}}{\sqrt{5}} \ln|x\sqrt{\frac{7}{5}}(1 + \frac{\sqrt{5}}{\sqrt{5}})| + C $$

$$ = \frac{7\sqrt{7}}{\sqrt{5}} \ln|x\sqrt{\frac{7}{5}}(1 + \sqrt{5})| + C $$