so which you at present have quintessential (2tan^2(theta)[2sec^2(theta) d(theta)/[2sec(theta)] This incorporates simplifying the sq. root section. So not extra sq. root (which replaced into the reason we did the trig substitution). Now an quintessential with a great number of sines and cosines. returned, based on the denominator, replace u=sin(theta), u=cos(theta), or u=tan(theta). This final one might undo the previous substitution, so it somewhat is no longer used right here. desire this enables!
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Using the Wolfram Integrator:
http://integrals.wolfram.com/index.jsp
One obtains
1/10 sec(2x)tan^4(2x) + 2/15 sec(2x)tan^2(2x) + 4/15 tan(2x)
Using the other poster's logic - you get (u^2 + 1)^3
Make the substitution u=tan(2x), du=2(sec(2x))^2dx.
and then rewrite your integral using the identity tan(x)^2+1=sec(x)^2.
sec(2x)^2=tan(2x)^2+1=u^2+1
sec(2x)^4=(tan(2x)^2+1)^2=(u^2+1)^2=u^4+2u^2+1
So then your integral becomes 0.5 $ (u^4+2u^2+1)du where $ is the integral sign. Now you can easily do it so do it and then undo the substitution.
so which you at present have quintessential (2tan^2(theta)[2sec^2(theta) d(theta)/[2sec(theta)] This incorporates simplifying the sq. root section. So not extra sq. root (which replaced into the reason we did the trig substitution). Now an quintessential with a great number of sines and cosines. returned, based on the denominator, replace u=sin(theta), u=cos(theta), or u=tan(theta). This final one might undo the previous substitution, so it somewhat is no longer used right here. desire this enables!