1
\begin{align*}
\langle H'_2 \rangle 
  &= \int_{0}^{b} |A|^2 sin^2(kr) \frac{-V_0r^4}{2b^4} dr\\
  &= \frac{(-V_0)}{2b^4}\int_{0}^{b} \frac{1-cos(2kr)}{2} r^4 dr \\
  &= \frac{(-V_0)}{2b^4}\frac{1}{2} 
     \left\lgroup \int_{0}^{b} r^4 dr 
     - \int_{0}^{b} r^4cos(2kr) dr \right\rgroup \\
  &= \frac{(-V_0)}{2b^4}\frac{1}{2} 
     \left\lgroup \frac{r^5}{5} - \frac{b^4sin(2kb)}{2k} 
     + \frac{1}{2k}\int_{0}^{b} 4r^3sin(2kr) dr \right\rgroup \\
  &= \frac{(-V_0)}{2b^4}\frac{1}{2} 
     \left\lgroup \frac{r^5}{5} - \frac{b^4sin(2kb)}{2k} 
     - \frac{4b^3(cos(2kb)-1)}{(2k)^2} 
     + \frac{4}{(2k)^2} \int_{0}^{b} 3r^2cos(2kr) dr \right\rgroup \\
  &= \frac{(-V_0)}{2b^4}\frac{1}{2} 
     \left\lgroup \frac{r^5}{5} - \frac{b^4sin(2kb)}{2k} 
     - \frac{(4b^3cos(2kb)-1)}{(2k)^2} + \frac{12b^2sin(2kb)}{(2k)^3} 
     - \frac{12}{(2k)^3} \int_{0}^{b} 2rsin(2kr) dr \right\rgroup\\       
  &= \frac{(-V_0)}{2b^4}\frac{1}{2} 
     \left\lgroup \frac{r^5}{5} - \frac{b^4sin(2kb)}{2k} 
     - \frac{(4b^3cos(2kb)-1)}{(2k)^2} 
     + \frac{12b^2sin(2kb)}{(2k)^3} \\
  &\quad+ \frac{24b(cos(2kb)-1)}{(2k)^4} - \frac{24}{(2k)^4} 
     \int_{0}^{b} cos(2kr) dr \right\rgroup
\end{align*}
  • 3
    Welcome! You cannot have a ` \\` between the \left\lgroup and \right\rgroup in the last line of your equation. – user121799 Jan 8 '18 at 13:04
  • 2
    You might be better off replacing \left and \right by \Bigl and \Bigr, which don't have this issue. (You may need to experiment to find the correct amount of \bigness.) By the way, I edited the title of the post. We don't do groping around here. – Harald Hanche-Olsen Jan 8 '18 at 13:16
  • And, if you’re interested in good math typography, please write \sin and \cos, not just sin and cos. Plus, consider inserting \, (“thinspace”) before each and every dr term. – Mico Jan 8 '18 at 13:29
3

You need \biggl and \biggr rather than \left and \right.

I'd prefer parentheses to \lgroup and \rgroup, though.

Here's a possibly better implementation, with slightly different alignment than proposed. I also fixed \sin and \cos instead of sin and cos; also added \, in front of dr. Finally, I changed r^5 into b^5.

The split happens at the left parenthesis, so the role of the second line is clearer; between the main parts I added some vertical space for better separation.

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{align*}
\langle H'_2 \rangle 
  &= \int_{0}^{b} |A|^2 \sin^2(kr) \frac{-V_0r^4}{2b^4} \,dr\\[1ex]
  &= \frac{(-V_0)}{2b^4}\int_{0}^{b} \frac{1-\cos(2kr)}{2} r^4 \,dr \\[2ex]
  &= \frac{(-V_0)}{2b^4}\frac{1}{2}
     \biggl\lgroup \int_{0}^{b} r^4 \,dr 
     - \int_{0}^{b} r^4\cos(2kr) \,dr \biggr\rgroup \\[2ex]
  &= \frac{(-V_0)}{2b^4}\frac{1}{2} 
     \biggl\lgroup \frac{b^5}{5} - \frac{b^4\sin(2kb)}{2k} 
     + \frac{1}{2k}\int_{0}^{b} 4r^3\sin(2kr) \,dr \biggr\rgroup \\[2ex]
  &= \frac{(-V_0)}{2b^4}\frac{1}{2}
     \biggl\lgroup\begin{aligned}[t]
     &\frac{b^5}{5} - \frac{b^4\sin(2kb)}{2k} - \frac{4b^3(\cos(2kb)-1)}{(2k)^2} \\
     &+ \frac{4}{(2k)^2} \int_{0}^{b} 3r^2\cos(2kr) \,dr \biggr\rgroup
     \end{aligned} \\[2ex]
  &= \frac{(-V_0)}{2b^4}\frac{1}{2} 
     \biggl\lgroup\begin{aligned}[t]
     & \frac{b^5}{5} - \frac{b^4\sin(2kb)}{2k} - \frac{(4b^3\cos(2kb)-1)}{(2k)^2} \\
     &+ \frac{12b^2\sin(2kb)}{(2k)^3} - \frac{12}{(2k)^3} \int_{0}^{b} 2r\sin(2kr) \,dr \biggr\rgroup
     \end{aligned} \\[2ex]
  &= \frac{(-V_0)}{2b^4}\frac{1}{2} 
     \biggl\lgroup\begin{aligned}[t]
     &\frac{b^5}{5} - \frac{b^4\sin(2kb)}{2k} 
     - \frac{(4b^3\cos(2kb)-1)}{(2k)^2} 
     + \frac{12b^2\sin(2kb)}{(2k)^3} \\
     &+ \frac{24b(\cos(2kb)-1)}{(2k)^4} - \frac{24}{(2k)^4} 
     \int_{0}^{b} \cos(2kr) \,dr \biggr\rgroup
     \end{aligned}
\end{align*}

\end{document}

enter image description here

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.