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0
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8
avatar+6881 

Still integrals, but harder than those integrals before.

 

1)\(\displaystyle \int^{1}_{0}\sqrt{1-\sqrt{x}}\;dx\)

2)\(\displaystyle \int^{1}_{0}\sqrt{\sqrt[3]{x}-\sqrt{x}}\;dx\)

3)\(\displaystyle \int^{1}_{0}(x^{2/5}-\sqrt[3]{x})^{4/7}\;dx\)

4)\(\displaystyle \int^{\pi/2}_{0}\cos^{15}{x}\;dx\)

5)\(\displaystyle \int^{\pi/2}_{0}\sin^{13}x\;dx\)

MaxWong  Nov 4, 2017
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8+0 Answers

 #1
avatar+18715 
+2

integrals

 

1)  \(\displaystyle \int \limits_{0}^{1}\sqrt{1-\sqrt{x}}\;dx\)

 

 For the integrand  \(\sqrt{1-\sqrt{x}} \) , substitute \(u = \sqrt{1-\sqrt{x}} \) or \(\sqrt{x}=1-u^2\) and

 \(\begin{array}{rcll} du &=& \frac12(1-\sqrt{x})^{\frac12-1} (-\frac12)x^{\frac12-1} \;dx \\ &=& -\dfrac14\cdot \dfrac{1}{\sqrt{1-\sqrt{x}}\sqrt{x}} \;dx \\ &=& -\dfrac14\cdot \dfrac{1}{\sqrt{1-\sqrt{x}}\sqrt{x}} \;dx \\ &=& -\dfrac14\cdot \dfrac{1}{u(1-u^2)} \;dx \\\\ \text{so } dx &=& -4u(1-u^2)\;du \end{array}\)

 

This gives a new lower bound \(u = \sqrt{1} = 1 \) and upper bound \(u = \sqrt{0} = 0\)
\(\begin{array}{rcll} &=& -4 \displaystyle \int \limits_{1}^{0}u\cdot u(1-u^2)\;du \\ &=& -4 \displaystyle \int \limits_{1}^{0}u^2(1-u^2)\;du \\ &=& -4 \displaystyle \int \limits_{1}^{0}u^2-u^4\;du \\ &=& -4\cdot \Big( \displaystyle \int \limits_{1}^{0}u^2 \;du-\displaystyle \int \limits_{1}^{0} u^4\;du \Big) \\ &=& -4\cdot \Big( \Big[ \frac{u^3}{3} \Big]_{1}^{0} -\Big[ \frac{u^5}{5} \Big]_{1}^{0} \Big) \\ &=& -4\cdot \Big( -\frac{1}{3} - \Big( -\frac{1}{5} \Big) \Big) \\ &=& -4\cdot \Big( \frac{1}{5} -\frac{1}{3} \Big) \\ &=& -4\cdot \left( \dfrac{-2}{15} \right) \\ &=& \dfrac{8}{15} \\ \end{array} \)

 

laugh

heureka  Nov 6, 2017
 #2
avatar+18715 
+2

integrals


2) \(\displaystyle \int \limits_{0}^{1}\sqrt{\sqrt[3]{x}-\sqrt{x}}\;dx\)

 

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{1}\sqrt{\sqrt[3]{x}-\sqrt{x}}\;dx \\ &=& \displaystyle \int \limits_{0}^{1}\sqrt{\sqrt[3]{x}\left(1-\frac{\sqrt{x}}{\sqrt[3]{x}} \right) }\;dx \\ &=& \displaystyle \int \limits_{0}^{1}\sqrt[6]{x}\sqrt{ 1-\sqrt[6]{x} }\;dx \\ \end{array} \)

 

For the integrand \(\sqrt[6]{x}\sqrt{ 1-\sqrt[6]{x}}\), substitute \(u = \sqrt[6]{x}\) or \(u^5 = x^{\frac{5}{6}}\) and

\(\begin{array}{rcll} du &=& \frac16 x^{\frac16-1} \;dx \\ &=& \frac16 x^{-\frac{5}{6}} \;dx \\ &=& \dfrac16\cdot \dfrac{1}{ x^{\frac{5}{6}}} \;dx \\ &=& \dfrac16\cdot \dfrac{1}{u^5} \;dx \\ \text{so } dx &=& 6u^5\;du \\ \end{array} \)

 

This gives a new lower bound \(u = \sqrt[6]{0} = 0\) and upper bound \(u = \sqrt[6]{1} = 1\)

\(\begin{array}{rcll} &=& \displaystyle \int \limits_{0}^{1}u\sqrt{1-u}\cdot 6u^5\;du \\ &=& 6\displaystyle \int \limits_{0}^{1}u^6\sqrt{1-u} \;du \\ \end{array} \)

 

For the integrand \(u^6\sqrt{1-u}\), substitute \(v = \sqrt{1-u}\)  or \(u = 1-v^2\) and

\(\begin{array}{rcll} dv &=& \dfrac{-1}{2\sqrt{1-u}} \;du \\ &=& \dfrac{-1}{2v} \;du \\ \text{so } du &=& -2v \;dv \\ \end{array}\)

 

This gives a new lower bound \(v = \sqrt{1} = 1\) and upper bound \(v = \sqrt{0} = 0\)

\(\begin{array}{rcll} &=& 6\displaystyle \int \limits_{1}^{0}(1-v^2)^6 v\cdot (-2)v\;dv \\ &=& -12\displaystyle \int \limits_{1}^{0}(1-v^2)^6 v^2\;dv \\ &=& -12 \displaystyle \int \limits_{1}^{0} (1-6v^2+15v^4-20v^6+15v^8-6v^{10}+v^{12} )v^2 \;dv \\ &=& -12 \displaystyle \int \limits_{1}^{0} ( v^2-6v^4+15v^6-20v^8+15v^{10}-6v^{12}+v^{14}) \;dv \\ &=& -12 \cdot \Big[ \frac{v^3}{3}-6\frac{v^5}{5}+15\frac{v^7}{7}-20\frac{v^9}{9}+15\frac{v^{11}}{11}-6\frac{v^{13}}{13}+\frac{v^{15}}{15} \Big]_{1}^{0} \\ &=& 12 \cdot\Big( \frac{1}{3}-\frac{6}{5}+\frac{15}{7}-\frac{20}{9}+\frac{15}{11}-\frac{6}{13}+\frac{1}{15} \Big) \\ &=& 12 \cdot\Big( \frac{675675-6\cdot405405+15\cdot289575-20\cdot225225+15\cdot184275-6\cdot 155925+135135}{2027025} \Big) \\ &=& \dfrac{552960}{2027025} \\ \\ &=& \dfrac{135\cdot 4096}{135\cdot 15015} \\ \\ &=& \dfrac{ 4096}{ 15015} \\ \end{array} \)

 

laugh

heureka  Nov 6, 2017
 #3
avatar+78755 
+2

4)

  

pi/ 2

 ∫     ( cos x)15   dx      =  

0

 

pi/ 2

 ∫   cos x   ( cos x)14   dx      = 

0

 

pi/ 2

 ∫   cos x   ( cos 2x)7   dx      = 

0

 

pi/ 2

 ∫   cos x  ( 1 - sin2 x)7   dx       

0

 

Let   u  =  sin x       →  du   = cos x dx

And  the new limits are :   sin (pi/2)  = 1      and  sin (0)   = 0

 

1

 ∫  ( 1 - u^2)7  du

0

 

1

∫  -u^14 + 7 u^12 - 21 u^10 + 35 u^8 - 35 u^6 + 21 u^4 - 7 u^2 + 1    du

0

 

                                                                                                                          1  

 [ - (1 /15)u15+(7/13)u13 - (21/11)u11+(35/9)u9 -(35/7)u7+(21/5)u5- (7/3)u3+ u ]

                                                                                                                          0

 

 [ -1/15   +  7/13  -  21/11  +  35/9   -  5  + 21/5  - 7/3  + 1 ]   

 

 

2048 / 6435 

 

 

 

cool cool cool     

CPhill  Nov 6, 2017
edited by CPhill  Nov 6, 2017
 #4
avatar+18715 
+2

integrals


4) \(\displaystyle \int^{\pi/2}_{0}\cos^{15}{x}\;dx\)

 

without substitution:

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \cos^{15}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x)\cos^{14}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x)\Big(\cos^2(x)\Big)^{7}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x) \Big(1-\sin^2(x) \Big)^{7}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x) \Big( \binom{7}{0}-\binom{7}{1}\sin^2(x) \\ && +\binom{7}{2}\sin^4(x)-\binom{7}{3}\sin^6(x)+\binom{7}{4}\sin^8(x) \\ && -\binom{7}{5}\sin^{10}(x)+\binom{7}{6}\sin^{12}(x)-\binom{7}{7}\sin^{14}(x) \Big)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\sin^0(x)\cos(x) -7\sin^2(x)\cos(x) \\ && +21\sin^4(x)\cos(x)-35\sin^6(x)\cos(x)+35\sin^8(x)\cos(x) \\ && -21\sin^{10}(x)\cos(x)+7\sin^{12}(x)\cos(x)-\sin^{14}(x)\cos(x) \Big)\;dx \\ \end{array}\)

 

Integration by parts

\(\begin{array}{rclrcl} u&=&\sin^n(x) & v &=& \sin(x) \\ u'&=&n\sin^{n-1}(x)\cos(x) & v' &=& \cos(x) \\ \end{array}\)

\(\begin{array}{|rcll|} \hline && \displaystyle \int \limits_{0}^{\pi/2} \underbrace{\sin^n(x)}_{u}\underbrace{\cos(x)}_{v'}\;dx \\ &=& \Big[ \underbrace{\sin^n(x)}_{u}\underbrace{\sin(x)}_{v} \Big]_{0}^{\pi/2} - \int \limits_{0}^{\pi/2} \underbrace{n \sin^{n-1}(x)\cos(x)}_{u'}\underbrace{\sin(x)}_{v}\;dx \\ &=& \Big[ \sin^{n+1}(x)\Big]_{0}^{\pi/2} - n \int \limits_{0}^{\pi/2}\sin^{n}(x)\cos(x)\;dx \\ \\ (n+1)\displaystyle \int \limits_{0}^{\pi/2} \sin^n(x)\cos(x)\;dx &=& \Big[ \sin^{n+1}(x)\Big]_{0}^{\pi/2} \\ \mathbf{\displaystyle \int \limits_{0}^{\pi/2} \sin^n(x)\cos(x)\;dx} & \mathbf{=} & \mathbf{\frac{\Big[\sin^{n+1}(x)\Big]_{0}^{\pi/2} } {n+1} = \frac{1}{n+1} }\\ \hline \end{array}\)

 

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\sin^0(x)\cos(x) -7\sin^2(x)\cos(x) \\ && +21\sin^4(x)\cos(x)-35\sin^6(x)\cos(x)+35\sin^8(x)\cos(x) \\ && -21\sin^{10}(x)\cos(x)+7\sin^{12}(x)\cos(x)-\sin^{14}(x)\cos(x) \Big)\;dx \\ &=& 1\cdot\frac{1}{1} - 7\cdot \frac{1}{3}+21\cdot \frac{1}{5} - 35\cdot \frac{1}{7}+35\cdot \frac{1}{9}- 21\cdot \frac{1}{11}+7\cdot \frac{1}{13} - 1\cdot \frac{1}{15} \\\\ &=& \mathbf{\dfrac{2048}{6435} } \\ \end{array}\)

 

laugh

heureka  Nov 6, 2017
 #5
avatar+18715 
+1

integrals


5) \(\displaystyle \int^{\pi/2}_{0}\sin^{13}x\;dx\)

 

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x)\sin^{12}(x)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x)\Big(\sin^2(x)\Big)^{6}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x) \Big(1-\cos^2(x) \Big)^{6}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x) \Big( \binom{6}{0}-\binom{6}{1}\cos^2(x) \\ && +\binom{6}{2}\cos^4(x)-\binom{6}{3}\cos^6(x)+\binom{6}{4}\cos^8(x) \\ && -\binom{6}{5}\cos^{10}(x)+\binom{6}{6}\cos^{12}(x) \Big)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\cos^0(x)\sin(x) -6\cos^2(x)\sin(x) \\ && +15\cos^4(x)\sin(x)-20\cos^6(x)\sin(x)+15\cos^8(x)\sin(x) \\ && -6\cos^{10}(x)\sin(x)+1\cos^{12}(x)\sin(x) \Big)\;dx \\ \end{array}\)

 

Integration by parts
\(\begin{array}{rclrcl} u&=&\cos^n(x) & v &=& -\cos(x) \\ u'&=&n\cos^{n-1}(x)\sin(x) & v' &=& \sin(x) \\ \end{array}\)

\(\small{ \begin{array}{|rcll|} \hline && \displaystyle \int \limits_{0}^{\pi/2} \underbrace{\cos^n(x)}_{u}\underbrace{\sin(x)}_{v'}\;dx \\ &=& \Big[ \underbrace{\cos^n(x)}_{u}(\underbrace{-\cos(x)}_{v}) \Big]_{0}^{\pi/2} - \int \limits_{0}^{\pi/2} \underbrace{-n \cos^{n-1}(x)\sin(x)}_{u'}(\underbrace{-\cos(x)}_{v})\;dx \\ &=& \Big[ -\cos^{n+1}(x)\Big]_{0}^{\pi/2} - n \int \limits_{0}^{\pi/2}\cos^{n}(x)\sin(x)\;dx \\ \\ (n+1)\displaystyle \int \limits_{0}^{\pi/2} \cos^n(x)\sin(x)\;dx &=& -\Big[ \cos^{n+1}(x)\Big]_{0}^{\pi/2} \\ \mathbf{\displaystyle \int \limits_{0}^{\pi/2} \cos^n(x)\sin(x)\;dx} & \mathbf{=} & \mathbf{\frac{-\Big[\cos^{n+1}(x)\Big]_{0}^{\pi/2} } {n+1} = \frac{1}{n+1} }\\ \hline \end{array} }\)

 

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\cos^0(x)\sin(x) -6\cos^2(x)\sin(x) \\ && +15\cos^4(x)\sin(x)-20\cos^6(x)\sin(x)+15\cos^8(x)\sin(x) \\ && -6\cos^{10}(x)\sin(x)+1\cos^{12}(x)\sin(x) \Big)\;dx \\ &=& 1\cdot\frac{1}{1} - 6\cdot \frac{1}{3}+15\cdot \frac{1}{5} - 20\cdot \frac{1}{7}+15\cdot \frac{1}{9}- 6\cdot \frac{1}{11}+1\cdot \frac{1}{13} \\\\ &=& \mathbf{\dfrac{1024}{3003} } \\ \end{array}\)

 

laugh

heureka  Nov 6, 2017
 #6
avatar+6881 
0

Well. I got quick ways for all of them.

 

Formula: \(\displaystyle\int^{1}_{0}x^p(1-x^r)^s\; dx = \dfrac{s}{p + rs + 1}B(\dfrac{p+1}{r},s)\)

Where \(B(x,y) = \dfrac{\Gamma(x) \cdot \Gamma(y)}{\Gamma(x+y)}\)

 

1)

 \(\displaystyle \int^{1}_{0}\sqrt{1-\sqrt{x}}dx\\ =\displaystyle \int^{1}_{0}x^{0}(1-x^{1/2})^{1/2}\; dx \\ =\dfrac{1/2}{0+1/2\cdot1/2 + 1}B(\dfrac{0+1}{1/2},1/2)\\ =\dfrac{2}{5}B(2,1/2)\\ =\dfrac{2}{5}\dfrac{\Gamma(2)\cdot\Gamma(1/2)}{\Gamma(2+1/2)}\\ =\dfrac{8}{15}\)

 

2)

\(\displaystyle\int^{1}_{0}\sqrt{\sqrt[3]{x}-\sqrt{x}}\; dx\\ =\displaystyle\int^{1}_{0}\sqrt{\sqrt[3]{x}(1-\sqrt[6]{x})}\;dx\\ =\displaystyle\int^{1}_{0}x^{1/6}(1-x^{1/6})^{1/2}\;dx\\ =\dfrac{1/2}{1/6+1+1/6\cdot1/2}B(\dfrac{1/6+1}{1/6},1/2)\;dx\\ =\dfrac{2}{5}B(7,1/2)\\ =\dfrac{4096}{15015}\)

3)

\(\displaystyle\int^{1}_{0}(x^{2/5}-\sqrt[3]{x})^{4/7}dx\\ =\displaystyle\int^{1}_{0}(x^{2/5})^{4/7}(1-x^{-1/15})^{4/7}dx\\ =\displaystyle\int^{1}_{0}x^{8/35}(1-x^{-1/15})^{4/7}dx\\ =\dfrac{\frac{4}{7}}{\frac{8}{35}+1+\frac{-1}{15}\cdot\frac{4}{7}}B(\dfrac{\frac{8}{35}+1}{\frac{-1}{15}},\dfrac{4}{7})\\ =\dfrac{12}{25}B(-\dfrac{129}{7},\dfrac{4}{7})\\ \approx -0.0626\)

4)

\(\displaystyle\int^{\pi/2}_{0}\cos^{15}xdx\\ =\displaystyle\int^{\pi/2}_{0}\cos x (1 - \sin^2 x)^7dx\\ u = \sin x\\ =\displaystyle\int^{1}_{0}(1-u^2)^7du\\ =\dfrac{7}{0+1+2\times7}B(\dfrac{0+1}{2},7)\\ =\dfrac{7}{15}B(1/2,7)\\ =\dfrac{2048}{6435}\)

5)

\(\displaystyle\int^{\pi/2}_{0}\sin^{13}xdx\\ =\displaystyle\int^{\pi/2}_{0}(-\sin x)(1-\cos^2x)^6\;dx\\ u = \cos x\\ =\displaystyle\int^{1}_{0}(1-u^2)^6du\\ =\dfrac{6}{0+1+2\times 6}B(\dfrac{0+1}{2},6)\\ =\dfrac{6}{13}B(\dfrac{1}{2},6)\\ =\dfrac{1024}{3003}\)

MaxWong  Nov 8, 2017
 #7
avatar+18715 
+2

With gamma and the beta function

 

Formula: \(\displaystyle \int \limits_{0}^{\pi/2} \cos^{2u-1}(x)\cdot \sin^{2v-1}(x) \;dx = \frac12 \cdot B(u,v), \qquad \text{Re } u>0, \quad \text{Re } v>0\)
Where \(B(x,y) = \dfrac{\Gamma(x) \cdot \Gamma(y)}{\Gamma(x+y)}, \qquad \text{Re } u>0, \quad \text{Re } v>0\)

 

4)\(\displaystyle \int \limits_{0}^{\pi/2} \cos^{15}(x)\;dx\)

 

\(\begin{array}{rr} \begin{array}{rr} \displaystyle \int \limits_{0}^{\pi/2} \cos^{15}(x)\cdot \;dx = \frac12 \cdot B(u,v) \\ \end{array}\\ \begin{array}{r|r} 2u-1 = 15 & 2v-1 = 0 \\ 2u = 16 & 2v = 1\\ u = 8 & v = \frac12 \\ \end{array} \end{array}\)

\(\begin{array}{rcll} \displaystyle \int \limits_{0}^{\pi/2} \cos^{15}(x)\cdot \;dx &=& \frac12 \cdot B(8,\frac12) \\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(8) \cdot \Gamma(\frac12)}{\Gamma(8+\frac12)} \quad & | \quad \Gamma(\frac12) = \sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(8) \cdot \sqrt{\pi} }{\Gamma(8+\frac12)} \quad & | \quad \Gamma(8) = 7! \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! \cdot \sqrt{\pi} }{\Gamma(8+\frac12)} \quad & | \quad \displaystyle \Gamma(8+\frac12) = \frac{(2\cdot 8)!}{8!4^8}\sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! \cdot \sqrt{\pi} }{\frac{(2\cdot 8)!}{8!4^8}\sqrt{\pi}} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! 8! 4^8}{16!} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 4^8}{ 9 \cdot 10 \cdot 11\cdot 12 \cdot 13 \cdot 14\cdot 15\cdot 16} \\\\ &=& \displaystyle \dfrac{165150720 }{518918400 } \\\\ &=& \displaystyle \dfrac{2048\cdot 80640 }{6435\cdot 80640 } \\\\ &=& \displaystyle \dfrac{2048 }{6435 } \\ \end{array}\)

 

 

5) \(\displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx\)

 

\(\begin{array}{rr} \begin{array}{rr} \displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx = \frac12 \cdot B(u,v) \\ \end{array}\\ \begin{array}{r|r} 2u-1 = 0 & 2v-1 = 13 \\ 2u = 1 & 2v = 14 \\ u = \frac12 & v = 7 \\ \end{array} \end{array}\)

\(\begin{array}{rcll} \displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx &=& \frac12 \cdot B(\frac12,7) \\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(\frac12) \cdot \Gamma(7)}{\Gamma(\frac12+7)} \quad & | \quad \Gamma(\frac12) = \sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot \Gamma(7)}{\Gamma(7+\frac12)} \quad & | \quad \Gamma(7) = 6! \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot 6!}{\Gamma(7+\frac12)} \quad & | \quad \displaystyle \Gamma(7+\frac12) = \frac{(2\cdot 7)!}{7!4^7}\sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot 6!}{\frac{(2\cdot 7)!}{7!4^7}\sqrt{\pi}} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{6! 7! 4^7}{14!} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6 \cdot 4^7}{ 8\cdot 9 \cdot 10 \cdot 11\cdot 12 \cdot 13 \cdot 14} \\\\ &=& \displaystyle \dfrac{5898240 }{17297280 } \\\\ &=& \displaystyle \dfrac{1024\cdot 5760 }{3003\cdot 5760 } \\\\ &=& \displaystyle \dfrac{1024 }{3003 } \\ \end{array}\)

 

laugh

heureka  Nov 9, 2017
edited by heureka  Nov 9, 2017
 #8
avatar+6881 
+1

Such a quick way!

The only formula used by me is:

\(\displaystyle \int^{1}_0 x^p(1-x^r)^sdx = \dfrac{s}{p+1+rs}B(\dfrac{p+1}{r},s)\)

MaxWong  Nov 12, 2017

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