Question

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

Answer 1

`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`