# Undoing Trigonometric Substitution

I've been learning integration in particular utilising trigonometric substitution. I understand all of it until the end, when I'm a bit confused about evaluating the integral in terms of the original function.

An example may best illustrate what I mean; $$\int x\sqrt{2x-x^2}dx$$ $x-1=\sin \theta, dx=\cos \theta d\theta$

$\int x\sqrt{2x-x^2}dx = \int (\sin \theta + 1)\cos ^2 \theta d\theta=-\frac{1}{3}\cos^3 \theta + \frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + C$

Which this particular textbook evaluates to; $-\frac{1}{3}(2x-x^2)^{\frac{3}{2}} + \frac{1}{2}\arcsin(\frac{x-1}{2}) + \frac{1}{4}\sin(2\arcsin(x-1)) + C$

Why don't they simplify the $\frac{1}{4}\sin(2\arcsin(x-1))$ by using double angle formulae and drawing the "triangle" (similar to this) to make it in terms of $x-1$?

I've found that with a variety of questions, some resources evaluate the integral by doing the "triangle" method and others just leave in in terms of inverse trigonometric functions.

Why is this? Is one answer more correct than the other? Is it something to do with domains (if so, what?)?

Thanks

• Laziness. And the point of the exercise is on the integration, not the back-substitution so to speak. There probably comes examples of what you are proposing later in the book =) – N3buchadnezzar Apr 22 '17 at 8:07
• @N3buchadnezzar do you need to be mindful of the domains of the trig substitution? – frog1944 Apr 22 '17 at 8:10
• In this context no. However, I would really suggest you read the following paper faculty.swosu.edu/michael.dougherty/book/chapter07.pdf. The first part shows the pitfall of using certain substitutions. The gist of it is that the substitution you use, need to have an continuous inverse. Especially 7.4 considers trigonometric substitutions. – N3buchadnezzar Apr 22 '17 at 10:01