Question

On the solution of the 13th exercise of the final exam, I don't get why sqrt((cos(pi*t^2 / 2))^2 + (sin(pi*t^2 / 2))^2) vanishes in the integral?

I do know the trig identity (cos(t))^2 + (sin(t))^2 = 1 but here we are integrating, and if we substitute pi*t^2 / 2 by a function u in order to apply the trig identity, we get du = pi*t dt or dt = du / (pi*t). And that subsitution isn't valid for the integral.

Here is the solution PDF : http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/final-exam/MIT18_01SCF10_finalsol.pdf

Answer 1

the trig identity holds inside the integral , so reduces to \( \int dt \)

however, you can do a change of variables, as you suggest:
\[ u= \frac{\pi t^2}{2} \\ t= \sqrt{\frac{2u}{\pi}}\]
\[ du = \pi t\ dt \\ dt= \frac{du}{\pi t} = \frac{1}{\sqrt{2\pi}}u^{-\frac{1}{2} } du\]
with limits t=0 -> u=0 and \(t=t_0\) -> \(u= \frac{\pi t_0^2}{2} \)
and the integral becomes
\[ \frac{1}{\sqrt{2\pi}}\int_0^\frac{\pi t_0^2}{2} (\cos^2(u)+\sin^2(u)) u^{-\frac{1}{2} }du \\ \frac{1}{\sqrt{2\pi}}\int_0^\frac{\pi t_0^2}{2} u^{-\frac{1}{2} }\ du

\]
if you integrate, you will get \(t_0\)

Answer 2

Oh ok that's right for the substitution it actually works! Does the substituion is necessary to justify that the trig identity works here?

It seems to me it shouldn't be correct to apply it like that without doing the substitution because the thing in the cos or the sin is not just a variable but another function. Or can we actually apply the trig identity no matter what's inside the cos and sin in an integral? Why this would be ok?

Answer 3

we can apply the trig identity. sin^2(stuff) + cos^2(stuff)
where stuff is exactly the same for both sin and cos, adds up to 1
it's (a bit) like
\[ x^2 - x^2 = 0 \]
adding up to zero
and that is also true inside an integral.

Answer 4

All right, thanks (once more) for the answer and the nlightenment!