hadi amiri 4 Messages 98 Reaction score 1 Thread starter Apr 25, 2010 #1 Evaluate [tex]\int\frac{arctan(x)dx}{(1+x^2)^\frac{3}{2}}[/tex]
arildno Science Advisor Homework Helper Gold Member Dearly Missed Messages 10,165 Reaction score 138 Apr 25, 2010 #2 Make the substitution: [tex]x=\tan(u),\to\frac{dx}{du}=\frac{1}{\cos^{2}u}[/tex] Thus, we get: [tex]dx=\frac{du}{\cos^{2}(u)}[/tex] and insertion in your integral yields: [tex]\int\frac{arctan(x)}{(1+x^{2})^{\frac{3}{2}}}=\int{u}\cos(u)du[/tex]
Make the substitution: [tex]x=\tan(u),\to\frac{dx}{du}=\frac{1}{\cos^{2}u}[/tex] Thus, we get: [tex]dx=\frac{du}{\cos^{2}(u)}[/tex] and insertion in your integral yields: [tex]\int\frac{arctan(x)}{(1+x^{2})^{\frac{3}{2}}}=\int{u}\cos(u)du[/tex]
hadi amiri 4 Messages 98 Reaction score 1 Apr 25, 2010 #3 your solution seems nice honestly i thought it is a hard one,becouse i picked it form "A coures of pure mathematics"
your solution seems nice honestly i thought it is a hard one,becouse i picked it form "A coures of pure mathematics"