1

It appears in my guide : the title exercise shows me the inicial substitution \begin{align} \int_{0}^{\pi/2}\cfrac{\sqrt{\sin x}\ dx }{\sqrt{\sin x}+\sqrt{\cos x}} & = \int_{0}^{\pi/2}\cfrac{\sqrt{\sin (\pi/2-y)}\ dy }{\sqrt{\sin (\pi/2-y)}+\sqrt{\cos (\pi/2-y)}} \\ & = -\int_{\pi/2}^{0}\cfrac{\sqrt{\cos y}\ dy }{\sqrt{\cos y}+\sqrt{\sin y}} \quad note \ that \sin (\pi/2-y)= \cos y, and \cos(\pi/2-y)=\sin y \\ & = \int_{0}^{\pi/2}\cfrac{\sqrt{\cos y}\ dy }{\sqrt{\cos y}+\sqrt{\sin y}} /* \cfrac{1}{\sqrt{\cos y}} \\ & = \int_{0}^{\pi/2}\cfrac{1}{1+\sqrt{\tan y}} \\ & = \int_{0}^{\pi/2}\cfrac{2u \ du}{\sec^2u(1+u)} \ I \ used \ u=\sqrt{\tan y} \\ & = \int_{0}^{\pi/2}\cfrac{2u \ du}{(\mathrm{u}^{3/2}+1)(1+u)} \ with \quad\sec^2y=\tan^2y-1 \\ \end{align} But I get stuck, I have tried parcial fractions, but it doesn´t work, any step wrong?

StubbornAtom
  • 17,052
Sebastian Fernandez
  • 123
  • Some steps wrong. You write $u=\sqrt{\tan x}$ when you mean $u=\sqrt{\tan y}$ ... and then the limits for $u$ are not $0$ to $\pi/2$. How did you get rid of $\sec^2 u$ ?? Try to re-write everything carefully. – GEdgar Apr 12 '19 at 00:16
  • Using $ \sec^2u \ as \tan^2u +1 $ , then by anterior substitution, we must get $\tan^2u+1 $ in terms of u, and we get $ \mathrm{u}^{3/2} $ – Sebastian Fernandez Apr 12 '19 at 00:39
  • https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 – lab bhattacharjee Apr 12 '19 at 01:18
  • Pretty clear, thanks. – Sebastian Fernandez Apr 12 '19 at 01:54
  • Of course $\tan^2 u + 1 \ne u^{3/2}$. Maybe $\tan^2 y + 1 = u^{3/2}$. This is what I mean by writing everything carefully. – GEdgar Apr 12 '19 at 11:22

1 Answers1

4

You don't need to make that substitution.

Watch closely. At no time do the fingers leave the hands.

$\begin{align} I &=\int_{0}^{\pi/2}\cfrac{\sqrt{\sin x}\ dx }{\sqrt{\sin x}+\sqrt{\cos x}} \\ & = \int_{0}^{\pi/2}\cfrac{\sqrt{\sin (\pi/2-y)}\ dy }{\sqrt{\sin (\pi/2-y)}+\sqrt{\cos (\pi/2-y)}} \\ & = -\int_{\pi/2}^{0}\cfrac{\sqrt{\cos y}\ dy }{\sqrt{\cos y}+\sqrt{\sin y}} \quad note \ that \sin (\pi/2-y)= \cos y, and \cos(\pi/2-y)=\sin y \\ \\ & = \int^{\pi/2}_{0}\cfrac{\sqrt{\cos y}\ dy }{\sqrt{\cos y}+\sqrt{\sin y}} \\ \end{align} $

Adding these two forms, $2I =\int^{\pi/2}_{0}\cfrac{\sqrt{\cos y}+\sqrt{\sin y}\ dy }{\sqrt{\cos y}+\sqrt{\sin y}} =\int^{\pi/2}_{0}dy =\dfrac{\pi}{2} $ so $I =\dfrac{\pi}{4} $.

Note this means that $\int_0^{\pi/2}\dfrac{f(\cos(x))}{f(\cos(x))+f(\sin(x))}dx =\int_0^{\pi/2}\dfrac{f(\sin(x))}{f(\cos(x))+f(\sin(x))}dx =\dfrac{\pi}{4} $.

The $\sqrt{}$ is a McGuffin.

marty cohen
  • 107,799