1. ## Double integration

double Integral (sin(theta)*x)dx dtheta

0 to 1/2pi and 1-cos(theta) to 0

2. Hi,

The question is:

$\displaystyle \int^{\frac{\pi}{2}} _ 0 \int ^{0} _{1-cos(\theta)}sin(\theta)\times {x}.dxd\theta$
$\displaystyle = \int ^{\frac{\pi}{2}} _ 0 \left ( \frac {x^2}{2} \times sin(\theta )\right ) |^{0} _{1-cos(\theta)}.d\theta$
$\displaystyle = \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 \left ( x^2\times sin(\theta )\right ) |^{0} _{1-cos(\theta)}.d\theta$
$\displaystyle = \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 \left (0\times sin(\theta ) - (1-cos(\theta))^2\times sin(\theta )\right ) .d\theta$
$\displaystyle = \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 \left ( - (1-2cos (\theta)+cos^2(\theta))\times sin(\theta )\right ) .d\theta$
$\displaystyle = \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 - sin(\theta )+2cos (\theta)sin(\theta )-cos^2(\theta)sin(\theta) .d\theta$
$\displaystyle = \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 - sin(\theta )+sin(2\theta )-\frac{1}{2}(cos(2\theta) +1)sin(\theta) .d\theta$
$\displaystyle = \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 - sin(\theta )+sin(2\theta )-\frac{1}{2}sin(\theta)cos(2\theta) -\frac{1}{2}sin(\theta) .d\theta$
$\displaystyle = \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 - sin(\theta )+sin(2\theta )-\frac{1}{4}(sin(\theta+2\theta)+sin(\theta-2\theta)) -\frac{1}{2}sin(\theta) .d\theta$
$\displaystyle = \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 - sin(\theta )+sin(2\theta )-\frac{1}{4}sin(3\theta)-\frac{1}{4}sin(-\theta) -\frac{1}{2}sin(\theta) .d\theta$
$\displaystyle = \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 - sin(\theta )+sin(2\theta )-\frac{1}{4}sin(3\theta)+\frac{1}{4}sin(\theta) -\frac{1}{2}sin(\theta) .d\theta$
$\displaystyle = \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 - \frac {5}{4} sin(\theta )+sin(2\theta )-\frac{1}{4}sin(3\theta) .d\theta$
$\displaystyle = \frac {1}{2}\left( - \frac{5}{4}cos(\theta )+\frac{1}{2}cos(2\theta )-\frac{1}{4}\times\frac{1}{3}cos(3\theta) \right )|^\frac{\pi}{2} _0$
$\displaystyle = \frac {1}{2}\left( -\frac{5}{4} cos(\theta )+\frac{1}{2}cos(2\theta )-\frac{1}{12}cos(3\theta) \right )|^\frac{\pi}{2} _0$

$\displaystyle =\frac {1}{2}(-\frac{1}{2} +\frac{5}{4} -\frac{1}{2} +\frac{1}{12})$
$\displaystyle =\frac{1}{6}$

the calculation is long forgive me if i made some mistakes.

3. It takes less time by reversing integration order.

4. Hello, cchyfly!

Another approach . . .

$\displaystyle \int^{\frac{\pi}{2}}_0 \underbrace{\int^{1-\cos\theta}_0\!\! x\cdot\sin\theta\,dx}\,d\theta$

$\displaystyle \int^{1-\cos\theta}_0\!\!x\cdot\sin\theta\,dx \;=\;\frac{1}{2}x^2\cdot\sin\theta\,\bigg]^{1-\cos\theta}_0 \;=\;\frac{1}{2}(1-\cos\theta)^2\sin\theta - \frac{1}{2}(0^2)\sin\theta$

. . $\displaystyle =\;\frac{1}{2}\left(1 - 2\cos\theta + \cos^2\!\theta\right)\sin\theta \;=\;\frac{1}{2}\left(\sin\theta - 2\sin\theta\cos\theta + \cos^2\!\theta\sin\theta\right)$

Then we have: .$\displaystyle \frac{1}{2}\int^{\frac{\pi}{2}}_0\left(\sin\theta - 2\sin\theta\cos\theta + \cos^2\theta\sin\theta\right)d\theta$ .$\displaystyle = \;\frac{1}{2}\bigg[-\cos\theta - \sin^2\!\theta -\frac{1}{3}\cos^3\!\theta\,\bigg]^{\frac{\pi}{2}}_0$

. . $\displaystyle = \;\frac{1}{2}\left(-\cos\frac{\pi}{2} - \sin^2\frac{\pi}{2} - \frac{1}{3}\cos^3\frac{\pi}{2}\right) - \frac{1}{2}\left(-\cos0 - \sin^20 - \frac{1}{3}\cos^30\right)$

. . $\displaystyle = \;\frac{1}{2}\left(\text{- }0 - 1 - 0\right) - \frac{1}{2}\left(\text{-}1 - 0 - \frac{1}{3}\right) \;=\;\frac{1}{2}\left(-1\right) - \frac{1}{2}\left(-\frac{4}{3}\right) \;=\;-\frac{1}{3} + \frac{2}{3}$

. . $\displaystyle = \;\frac{1}{6}$

5. Originally Posted by cchyfly
double Integral (sin(theta)*x)dx dtheta

0 to 1/2pi and 1-cos(theta) to 0
\displaystyle \begin{aligned} \int_{0}^{\pi /2}\!{\int_{1-\cos \theta }^{0}{x\sin \theta \,dx}\,d\theta }&=-\int_{0}^{1}\!{\int_{\arccos (1-x)}^{\pi /2}{x\sin \theta \,d\theta }\,dx} \\ & =\int_{0}^{1}{\left( x^{2}-x \right)\,dx} \\ & =-\frac{1}{6}. \\ \end{aligned}