Did a great job, relevant after 13 years. Keep it up !!
Code.with.Tanmay
God bless your soul - been staring at this problem for an hour.
morganolivianewton
What an elegant solution! My answer sheet would be filled with scribbles if a question like this comes out in exam!
acash
Fixed: First check that (1 + u^2)(u^4 - u^2 + 1) = u^6 + 1; knowing this is the clever part. Now d(tan(x)) = sec^2(x)dx = (1+tan^2(x))dx. Substitute u=tan(x) to take integral of u^4 - u^2 + 1, and finish. The key is 1+tan^2(x)=sec^2(x). The way to remember this is to divide the equation cos^2(x) + sin^(x) = 1 by cos^2(x).
MathDoctorBob
It gets to the same answer. That first factorization is pretty slick and not obvious.
MathDoctorBob
Thanks! Been there, done that. :) Elegance comes with a lot of practice. - Bob
MathDoctorBob
Same methodology, but applied to cot^6(x). Changes include cot^2(x) + 1 = csc^2(x) and cot(x)' = -csc^2(x)
MathDoctorBob
The hard work for tan^6(x) is done. The integral of 1 is x.
MathDoctorBob
the answer i got from my teacher is (1+tan^2(x)) (tan^4(x)-tan^2(x)+1)
then. (tan^4(x)-tan^2(x)+1)d(tanx)
and continue from there.
Can i do it separately? like integrate 1 and integrate tan^6(x) and still get the same answer?
momo
is it reduction method or other method ?
pujitakrishnan
hey man help me please, I can't understand how you eliminate the sec^2(x)
MrJuanhok
how do i do it if its a Integral of 1 + tan^6(x)?
momo
thanks. Would be great if you could solve it that way for me. I tried but couldn't get it. lol. thanks in advance.
