changing the order of integration

**sawawak** · April 16th 2009, 03:10 AM

I need to get sin1(1-cos1) for the answer but I keep getting wrong answer. Please someone help me !!

**Soroban** · April 16th 2009, 04:08 AM

Hello, sawawak!

I too can get only so far . . .

$\int^1_0\int^1_x y^{-1}\sin y\cos\left(\frac{x}{y}\right)\,dy\,dx$

I need to get: . $\sin1(1-\cos1)$ for the answer.

The limits are: . $\begin{Bmatrix}y = 1 \\ y=x\end{Bmatrix}\quad\begin{Bmatrix}x=1 \\ x=0\end{Bmatrix}$

The region looks like this:

Code:

        |
       1+ - - - - - *
        |:::::::::* :
        |:::::::*   :
        |:::::*     :
        |:::*       :
        |:*         :
      - * - - - - - + - -
        |           1

Reversing the order of integration, we have: . $\begin{Bmatrix}x=y\\ x=0 \end{Bmatrix}\quad\begin{Bmatrix}y=1 \\ y=0 \end{Bmatrix}$

And we have: . $\int^1_0\underbrace{\int^y_0 y^{-1}\cdot\sin y\cdot\cos\left(\frac{x}{y}\right)\,dx}\,dy$

. . . . . . . . . . . . $y^{-1}\cdot\sin y\cdot\frac{1}{y}\cdot\sin\left(\frac{x}{y}\right) \,\bigg]^y_0$

. . . . . . . . . $= \;\bigg[y^{-1}\sin y\cdot\frac{1}{y}\cdot\sin\left(\frac{y}{y}\right) \bigg] - \bigg[y^{-1}\cdot\sin y \cdot\frac{1}{y}\cdot\sin\left(\frac{0}{y}\right)\ bigg]$

. . . . . . . . . $= \;\frac{1}{y}\cdot\sin y\cdot\frac{1}{y}\cdot\sin(1)$

. . . . . . . . . $= \;\sin(1)\,\frac{\sin y}{y^2}$

Now we must integrate: . $\sin(1)\int^1_0 \frac{\sin y}{y^2}\,dy$

Did I make some egregious blunders?

**Danny** · April 16th 2009, 04:31 AM

The $\frac{1}{y}$ get absorbed on integration. I'll pick it up here

$\int^1_0\int^y_0 y^{-1}\cdot\sin y\cdot\cos\left(\frac{x}{y}\right)\,dx\,dy = \int_0^1 \sin y \left. \sin \frac{x}{y} \right|_0^y\, dy = \int_0^1 \sin y \sin 1\, dy$ $= - \sin 1 \cos y |_0^1 = \sin 1 ( 1 - \cos 1)$

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