1
$\begingroup$

I want to make the substition $u = \cos^2\left(\dfrac{\pi t}{2}\right)$

Now I understand that $du = -\pi \sin\left(\dfrac{\pi t}{2}\right)\cos\left(\dfrac{\pi t}{2}\right)dt$

But then my professor wrote $\cos\left(\dfrac{\pi t}{2}\right)~dt = -\dfrac{1}{\pi}du \sin\left(\dfrac{\pi t}{2}\right)$ Which I dont understand.

Shouldn't it be $\cos\left(\dfrac{\pi t}{2}\right)~dt = -\dfrac{1}{\pi}~du \dfrac{1}{\sin\left(\dfrac{\pi t}{2}\right)}$?

  • 1
    you are correct! – 2017-01-05
  • 1
    @ArnaldoNascimento Yes but the thing is, he uses the fact that $sin(\pi t/2) = (1-u)^{1/2}$ to then solve an integral – 2017-01-05
  • 0
    He cans do that but using in your expression. – 2017-01-05
  • 0
    Could you provide the original integral in which you are making the substitution? – 2017-01-05
  • 0
    @amWhy The original integral in which he makes the substitution is: $\int_0 ^1 (sin(\pi t/2)^ncos(\pi t /2)dt$ – 2017-01-05
  • 0
    @SylvesterStallone No this is not the right integral. – 2017-01-05
  • 0
    For that integral you should make $u=(\sin(\pi t/2))^{n+1}$. – 2017-01-05

1 Answers 1

0

Just a suggestion for solve the integral:

$$\int_0 ^1 \left[\sin\left(\frac{\pi t}{2}\right)\right]^{n}\cdot \cos\left(\frac{\pi t}{2}\right)dt$$

a better substitution is

$$u=\left[\sin\left(\frac{\pi t}{2}\right)\right]^{n+1} \rightarrow du=\frac{\pi(n+1)}{2}\left[\sin\left(\frac{\pi t}{2}\right)\right]^{n}\cdot \cos\left(\frac{\pi t}{2}\right)dt$$