1
$\begingroup$

I've been trying to integrate $$\int\frac{dx}{(x-a)(x-b)}$$

By using the substitution $$x=a \cos^2 \theta + b \sin^2 \theta$$

The only problem here is I arrived at the result $$\frac{2}{a-b} \ln |\csc 2\theta - \cot 2\theta|+c$$

and I having trouble on how to substitute from $\theta$ to $x$.

$\endgroup$
  • $\begingroup$ Update: I think I've found the solution: It's the Pythagorean identity $\endgroup$ – Loo Soo Yong Feb 27 '19 at 18:49
  • $\begingroup$ Do you have that $a$ and $b$ are both positive real numbers? Or is it allowed that one or both are negative? $\endgroup$ – Clayton Feb 27 '19 at 19:06
  • $\begingroup$ I believe that both are positive real numbers $\endgroup$ – Loo Soo Yong Feb 27 '19 at 19:11
  • $\begingroup$ Fantastic substitution btw! $\endgroup$ – user150203 Mar 1 '19 at 9:58
1
$\begingroup$

You can use a Weierstrass substitution: $$t:=\tan\theta\implies\csc 2\theta-\cot 2\theta=\frac{1+t^2}{2t}-\frac{1-t^2}{2t}=t=\sqrt{\frac{x-a}{b-x}}$$since$$x-a=(b-a)\sin^2\theta,\,b-x=(b-a)\cos^2\theta.$$

$\endgroup$
1
$\begingroup$

Take the middle of the two roots $x_0=\dfrac{a+b}2$ and make a substitution $t=x-x_0$

Then $\displaystyle\int\dfrac{\mathop{dx}}{(x-a)(x-b)}=\int\dfrac{\mathop{dt}}{\left(t^2-(\frac{a-b}2)^2\right)}=-\dfrac 2{a-b}\tanh^{-1}\left(\dfrac{2t}{a-b}\right)+C\quad$ for $x\in[a,b]$

Note: for $x$ outside $[a,b]$ use the form in $\operatorname{cotanh}^{-1}$ instead.


Rem: using the formula $\tanh^{-1}(x)=\dfrac 12\ln\left(\dfrac{1+x}{1-x}\right)$, I let you convince yourself it is the same as the answer given by GNU or Clayton.

In a doubt, you can look at this post for a similar approach, where I calculated in detail the equality with other forms of the result.

Integral of product of two inverse polynomials

$\endgroup$
1
$\begingroup$

This is not an answer, but I am posting an alternate method for fun. This algebraic approach should be simpler than the trigonometric substitution, and thus it's useful in exams/tests in which time is limited.

Observe that the integrand is a rational function which can be decomposed into two simple .

$$\frac{1}{(x-a)(x-b)} = \frac{1}{a-b} \left[ \frac{1}{x-a} - \frac{1}{x-b}\right]$$

This gives

\begin{aligned} \int\frac{dx}{(x-a)(x-b)} =& \frac{1}{a-b} \left[ \int\frac{dx}{x-a} - \int\frac{dx}{x-b}\right] \\ =& \frac{1}{a-b} (\ln|x-a| - \ln|x-b|) + C. \end{aligned}

$\endgroup$
1
$\begingroup$

As stated in the comments, we assume $a$ and $b$ are positive real numbers. In this case, without loss of generality, we may assume $a\leq b$. This gives $$x=a\cos^2(\theta)+b\sin^2(\theta)=a\cos^2\theta+a\sin^2\theta+(b-a)\sin^2\theta=a+(b-a)\sin^2\theta.$$This allows us to solve for $\theta$ so that we find $$\theta=\arcsin\sqrt{\frac{x-a}{b-a}\,}.$$Plugging into $\csc(2\theta)$ and $\cot(2\theta)$, applying double-angle formulas, and reducing the trigonometry (first using algebra to reduce to $\tan\theta$ and then using triangles), we find that $$\int\frac{1}{(x-a)(x-b)}\,dx = \frac{1}{a-b}\left(\ln|x-a|-\ln|x-b|\right)+C.$$

$\endgroup$
  • $\begingroup$ The sign of the coefficient on the left should be inverted. Please see my answer as well as Wolfram Alpha for reference. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 8 '19 at 9:23
  • $\begingroup$ It was a typo, @GNUSupporter8964民主女神地下教會: thanks. I used Wolfram to verify as well :P Also, I can't speak for others, but I'd suggest definitely feel free to correct typos like that in the future :) $\endgroup$ – Clayton Mar 8 '19 at 13:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.