0

This integral is given without proof/detail in the textbook Modern Compressible Flow by John Anderson. The ratio of specific heats, $\gamma$, is a positive and constant real number.

$\int_{M_1}^{M_2}\frac{2}{\gamma M^2}(1-M^2)\left[1+\frac{1}{2}(\gamma-1)M^2\right]^{-1}\frac{dM}{M}=\left[-\frac{1}{\gamma M^2}-\frac{\gamma+1}{2\gamma}\ln\left(\frac{M^2}{1+\frac{\gamma-1}{2}M^2}\right)\right]_{M_1}^{M_2}$

But I can't figure out how to integrate it. My initial thought is that integration by parts was used, but I haven't had any luck. I also tried a few different substitutions such as $u=1+\frac{1}{2}(\gamma-1)M^2$, so $du=(\gamma -1)MdM$, but that clearly doesn't work.

How can I integrate the given expression?

nwsteg
  • 329
  • 1
    If you're given the indefinite integral and you want to verify it, it's usually better to differentiate the result and verify you get the integrand. Doing so will usually also tell you the steps necessary to integrate the expression; chain rule indicates a substitution, and product rule suggests integration by parts. – Theo Bendit Oct 18 '22 at 01:01

1 Answers1

3

The integral is just a rational function in $M$; factoring out the constants and writing $a = (\gamma - 1)/2$, you can reduce the integral to $$\int_{M_1}^{M_2} \frac{(1 - M^2)}{(1 + aM^2)M^3} \, dM.$$ I'm sure you can proceed from here.

Edit: to simplify the integral, you can make the substitution $u = M^2$ and write $$\int \frac{1 - M^2}{(1 + aM^2)M^3} \, dM = \frac 12 \int \frac{1 - u}{(1 + au)u^2} \, du$$ to simplify the partial fraction process.

Emory Sun
  • 1,536