T tutor01 Member Joined Jul 21, 2005 Messages 46 Gender Male HSC N/A May 9, 2017 #1 I am trying to find the integral of dx/(3+4sin(2x)). Can anyone help? Thanks!

S si2136 Well-Known Member Joined Jul 19, 2014 Messages 1,383 Gender Undisclosed HSC N/A May 9, 2017 #2 Use the t formula

T tutor01 Member Joined Jul 21, 2005 Messages 46 Gender Male HSC N/A May 9, 2017 #3 was hoping for a solution. t=tanx and from there. Is that best? Thanks.

pikachu975 Premium Member Joined May 31, 2015 Messages 2,588 Location NSW Gender Male HSC 2017 May 9, 2017 #4 tutor01 said: was hoping for a solution. t=tanx and from there. Is that best? Thanks. Click to expand... I reckon splitting sin2x into 2sinxcosx could work too. t = tan(x/2) x = 2tan^-1 (t) dx = 2dt/(1+t^2) Integral of 1/(3+8(2t/1+t^2)(1-t^2 /1+t^2)) * 2dt/(1+t^2) Go from there, just simplify then integrate

tutor01 said: was hoping for a solution. t=tanx and from there. Is that best? Thanks. Click to expand... I reckon splitting sin2x into 2sinxcosx could work too. t = tan(x/2) x = 2tan^-1 (t) dx = 2dt/(1+t^2) Integral of 1/(3+8(2t/1+t^2)(1-t^2 /1+t^2)) * 2dt/(1+t^2) Go from there, just simplify then integrate

integral95 Well-Known Member Joined Dec 16, 2012 Messages 780 Gender Male HSC 2013 May 9, 2017 #5 Yes I believe t = tan x would be a better substitution in this integral.

integral95 Well-Known Member Joined Dec 16, 2012 Messages 780 Gender Male HSC 2013 May 9, 2017 #6 Wolfram alpha gives an inverse hyperbolic tan function. So have fun.