Integration of Rational Functions with Trigonometric Substitutions

[hms.tech]

What is the procedure to integrate this kind of a fraction, i am guessing some trigonometric
identity will apply but i am not quite sure how


∫[1/(a-bx^2)] dx

can someone provide me a link which would help me learn about this kind of integration.

Any help is Appreciated

 

[DonAntonio]

What is the procedure to integrate this kind of a fraction, i am guessing some trigonometric
identity will apply but i am not quite sure how


∫[1/(a-bx^2)] dx

can someone provide me a link which would help me learn about this kind of integration.

Any help is Appreciated

Google "partial fractions integration". In your case,
$$\frac{1}{a-bx^2}=\frac{1}{(\sqrt a-\sqrt bx)(\sqrt a+\sqrt bx)}$$
iff $\,a,b\geq 0\,$

DonAntonio

 

[mathman]

The above works as long as a and b have the same sign - if both are negative, replace by minus absolute value. If they are of opposite sign, then you end up with something involving an arctan.

 

[johnqwertyful]

Can't you just pull out a ${b}$, giving $\frac{1}{{b}(\frac{a}{{b}}-x^{2})}$

Then use a simple inverse sin formula?

 

[Millennial]

You can, but that does not mean you have to. It depends on what solution is being asked from you, because for example you are asked to solve the problem with partial fractions and you try a sine substitution, you will have to break up your answer into two natural logarithms later on. For definite integrals though, this often is not a problem.

 

[DonAntonio]

Can't you just pull out a ${b}$, giving $\frac{1}{{b}(\frac{a}{{b}}-x^{2})}$

Then use a simple inverse sin formula?

Inverse sine formula? For that one needs $\,(\sqrt{1-x^2})^{-1}\,$ and we have no square roots here.

DonAntonio

 

[Millennial]

DonAntonio, it is still applicable. Let $\sin(t)=x$. Then we have that $\displaystyle \int \frac{dt}{\cos(t)}=\int \sec(t)dt$ which has a closed-form solution.

 

[johnqwertyful]

Inverse sine formula? For that one needs $\,(\sqrt{1-x^2})^{-1}\,$ and we have no square roots here.

DonAntonio

Yeah, you need a square root for inverse sin, I knew that.
But still, just do a trig sub.

 

[DonAntonio]

DonAntonio, it is still applicable. Let $\sin(t)=x$. Then we have that $\displaystyle \int \frac{dt}{\cos(t)}=\int \sec(t)dt$ which has a closed-form solution.

Yes...so? I was referring to other poster's idea of using something about the inverse sine, or $\,\arcsin\,$ , which I

can't see how it can appear here more or less "naturally". Of course, one can alwasy make up substitutions as to make

almost any function pop up, but this is not what I meant.

DonAntonio