# Math Help - How to evaluate this integral by hand?

1. ## How to evaluate this integral by hand?

Hello everyone,
$$\int\limits_0^{\sqrt 2 h} {\frac{\tau }{{\sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}d} \tau$$
I want to evaluate the above integral by hand. It is from a question in Griffiths Introduction to Electrodynamics. Can you explain how to do step by step?

P.S: I did it with Wolframalpha but failed to understand it since it gave the desired answer for restricted values of x and h values. What does that mean by the way?

2. ## Re: How to evaluate this integral by hand?

Originally Posted by JohnDoe
Hello everyone,
$$\int\limits_0^{\sqrt 2 h} {\frac{\tau }{{\sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}d} \tau$$
I have not done the integral, but I suspect it involves:
$r^2 + h^2 - \sqrt 2 hr = \left( {r - \frac{h}{{\sqrt 2 }}} \right)^2 + \frac{{h^2 }}{2}~.$

+h²/2 only

4. ## Re: How to evaluate this integral by hand?

Yes thanks for your help but I am stuck after these step, I cannot find the integral I might be doing something wrong here is my work:
$\frac{{\sigma 2\pi }}{{4\pi {\varepsilon _0}\sqrt 2 }}\int\limits_0^{\sqrt 2 h} {\frac{{\tau d\tau }}{{\sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}} \\
{\tau ^2} + {h^2} - \sqrt 2 h\tau = {(\tau - \frac{h}{{\sqrt 2 }})^2} + \frac{{{h^2}}}{2}\\
u = (\tau - \frac{h}{{\sqrt 2 }})\\
\frac{\sigma }{{\pi {\varepsilon _0}2\sqrt 2 }}\int\limits_a^b {\frac{{u + \frac{h}{{\sqrt 2 }}du}}{{\sqrt {{u^2} + \frac{{{h^2}}}{2}} }}} \\$

$\frac{h}{{\sqrt 2 }}\tan \theta = u\\
\frac{\sigma }{{\pi {\varepsilon _0}2\sqrt 2 }}\int\limits_c^d {\frac{{\frac{h}{{\sqrt 2 }}(1 + \tan \theta )\frac{h}{{\sqrt 2 }}{{\sec }^2}\theta d\theta }}{{\frac{h}{{\sqrt 2 }}\sec \theta }}} \\
\frac{\sigma }{{\pi {\varepsilon _0}4}}\int\limits_c^d {(\sec \theta + \sec \theta \tan \theta )d\theta } = \frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt {2{u^2} + {h^2}} }}{h} + \frac{{\sqrt 2 u}}{h}) + \frac{{\sqrt {2{u^2} + {h^2}} }}{h}]\\
$

$\frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt {2{{(\tau - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h} + \frac{{\sqrt 2 (\tau - \frac{h}{{\sqrt 2 }})}}{h}) + \frac{{\sqrt {2{{(\tau - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h}]_0^{\sqrt 2 h}\\
\frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } + \sqrt 2 \tau - h}}{h}) + \frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}{h}]_0^{\sqrt 2 h}$

Edit:I have added the original solution I think something is wrong there with the natural logarithm part.

5. ## Re: How to evaluate this integral by hand?

try using the substitution $u=\left (\tau-\frac{h}{\sqrt{2}} \right )^2$

6. ## Re: How to evaluate this integral by hand?

That the fourth line in the attached image contains a sum is possibly indicative of integration by parts having been used.

CB

7. ## Re: How to evaluate this integral by hand?

Originally Posted by JohnDoe
Yes thanks for your help but I am stuck after these step, I cannot find the integral I might be doing something wrong here is my work:
$\frac{{\sigma 2\pi }}{{4\pi {\varepsilon _0}\sqrt 2 }}\int\limits_0^{\sqrt 2 h} {\frac{{\tau d\tau }}{{\sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}} \\
{\tau ^2} + {h^2} - \sqrt 2 h\tau = {(\tau - \frac{h}{{\sqrt 2 }})^2} + \frac{{{h^2}}}{2}\\
u = (\tau - \frac{h}{{\sqrt 2 }})\\
\frac{\sigma }{{\pi {\varepsilon _0}2\sqrt 2 }}\int\limits_a^b {\frac{{u + \frac{h}{{\sqrt 2 }}du}}{{\sqrt {{u^2} + \frac{{{h^2}}}{2}} }}} \\$
No more $\pi$

$\frac{\sigma }{{{\varepsilon _0}2\sqrt 2 }}\int\limits_a^b {\frac{{u + \frac{h}{{\sqrt 2 }}du}}{{\sqrt {{u^2} + \frac{{{h^2}}}{2}} }}}$

Originally Posted by JohnDoe
$\frac{h}{{\sqrt 2 }}\tan \theta = u\\
\frac{\sigma }{{\pi {\varepsilon _0}2\sqrt 2 }}\int\limits_c^d {\frac{{\frac{h}{{\sqrt 2 }}(1 + \tan \theta )\frac{h}{{\sqrt 2 }}{{\sec }^2}\theta d\theta }}{{\frac{h}{{\sqrt 2 }}\sec \theta }}}
$
$\frac{\sigma }{{{\varepsilon _0}2\sqrt 2 }}\int\limits_c^d {\frac{{\frac{h}{{\sqrt 2 }}(1 + \tan \theta )\frac{h}{{\sqrt 2 }}{{\sec }^2}\theta d\theta }}{{\frac{h}{{\sqrt 2 }}\sec \theta }}}$

Originally Posted by JohnDoe
$\frac{\sigma }{{\pi {\varepsilon _0}4}}\int\limits_c^d {(\sec \theta + \sec \theta \tan \theta )d\theta } = \frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt {2{u^2} + {h^2}} }}{h} + \frac{{\sqrt 2 u}}{h}) + \frac{{\sqrt {2{u^2} + {h^2}} }}{h}]\\
$
h is missing at the numerator

$\frac{\sigma h}{{{\varepsilon _0}4}}\int\limits_c^d {(\sec \theta + \sec \theta \tan \theta )d\theta } = \frac{\sigma h}{{{\varepsilon _0}4}} \left[\ln (\frac{{\sqrt {2{u^2} + {h^2}} }}{h} + \frac{{\sqrt 2 u}}{h}) + \frac{{\sqrt {2{u^2} + {h^2}} }}{h} \right]$

Originally Posted by JohnDoe
$\frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt {2{{(\tau - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h} + \frac{{\sqrt 2 (\tau - \frac{h}{{\sqrt 2 }})}}{h}) + \frac{{\sqrt {2{{(\tau - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h}]_0^{\sqrt 2 h}\\
\frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } + \sqrt 2 \tau - h}}{h}) + \frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}{h}]_0^{\sqrt 2 h}$
$\frac{\sigma h}{{{\varepsilon _0}4}} \left[\ln (\frac{{\sqrt {2{{(\tau - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h} + \frac{{\sqrt 2 (\tau - \frac{h}{{\sqrt 2 }})}}{h}) + \frac{{\sqrt {2{{(\tau - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h} \right]_0^{\sqrt 2 h}\\
\frac{\sigma h}{{{\varepsilon _0}4}} \left[\ln (\frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } + \sqrt 2 \tau - h}}{h}) + \frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}{h} \right]_0^{\sqrt 2 h}$

Your calculation is correct, you just need to finish it to finally find $\frac{\sigma h}{{{\varepsilon _0}2}} \ln{\left(1+\sqrt 2 \right)}$

8. ## Re: How to evaluate this integral by hand?

Thanks for your help, but I am still unable to understand the fourth line in the picture I have attached. Because my solution expression before the evaluation of the integral does not match the solution printed on the manual. I got h in the demoninator inside ln, whereas the solution manual does not have any h in the denominator inside ln function. What may be the reason behind that? Is it possible to reach the answer using different indefinite integrals? Can someone provide the skipped steps?

9. ## Re: How to evaluate this integral by hand?

Originally Posted by JohnDoe
I got h in the demoninator inside ln, whereas the solution manual does not have any h in the denominator inside ln function.
It does not matter because

$\left[\ln \left(\frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } + \sqrt 2 \tau - h}}{h}\right)\right]_0^{\sqrt 2 h} = \left[\ln \left({\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } + \sqrt 2 \tau - h}\right) - \ln(h)\right]_0^{\sqrt 2 h}$

$\ln(h)$ is a constant that cancels out

Therefore

$\left[\ln \left(\frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } + \sqrt 2 \tau - h}}{h}\right)\right]_0^{\sqrt 2 h} = \left[\ln \left({\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } + \sqrt 2 \tau - h}\right)\right]_0^{\sqrt 2 h}$