How do you integrate #int x /sqrt( 16 - x^4 )dx# using trigonometric substitution?

Question

7280 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-21T18:07:46

you just need to do a normal substitution

#u = 1/4x^2#
#du = 1/2xdx#

#2int1/2x/sqrt(16-x^4)dx#

#2int1/sqrt(16-16u^2)du#

#1/2int1/sqrt(1-u^2)du#

as you know, it's the derivate of #arcsin(u)#

so

#= 1/2[arcsin(u)]+C#

#= 1/2[arcsin(x^2)]+C#

OF COURSE You can do it with #u = sin(t)#

#du = cos(t) dt#

and you have

#1/2intcos(t)/cos(t)dt#

#1/2intdt#

#1/2[t]#

the fact is #t = arcsin(u)# ...