How do you integrate #int t^5/sqrt(t^2+2)# by trigonometric substitution?

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mason m Share
Nov 23, 2016

Answer:

#(sqrt(t^2+2)(3t^4-8t^2+32))/15+C#

Explanation:

#I=intt^5/sqrt(t^2+2)dt#

Let #t=sqrt2tantheta#. This implies that #dt=sqrt2sec^2thetad theta#. Plugging these in yields

#I=int(sqrt2tantheta)^5/sqrt(2tan^2theta+2)(sqrt2sec^2thetad theta)#

#I=(sqrt2)^5int(tan^5thetasec^2theta)/sqrt(tan^2theta+1)d theta#

Since #tan^2theta+1=sec^2theta#:

#I=4sqrt2inttan^5thetasecthetad theta#

#I=4sqrt2inttan^4theta(tanthetasectheta)d theta#

#I=4sqrt2int(sec^2theta-1)^2(secthetatantheta)d theta#

Letting #u=sectheta# so #du=secthetatanthetad theta#

#I=4sqrt2int(u^2-1)^2du=4sqrt2int(u^4-2u^2+1)du#

#I=4sqrt2(u^5/5-2/3u^3+u)#

#I=4sqrt2sectheta(sec^4theta/5-2/3sec^2theta+1)#

Our original substitution was #tantheta=t/sqrt2#. Thus #sectheta=sqrt(tan^2theta+1)=sqrt(t^2/2+1)=1/sqrt2sqrt(t^2+2)#.

#I=4sqrt2(1/sqrt2sqrt(t^2+2))(1/5(sqrt((t^2+2)/2))^4-2/3(sqrt((t^2+2)/2))^2+1)#

#I=4sqrt(t^2+2)(1/5((t^2+2)/2)^2-2/3((t^2+2)/2)+1)#

#I=4sqrt(t^2+2)((t^4+4t^2+4)/20-(t^2+2)/3+1)#

#I=4sqrt(t^2+2)((3t^4+12t^2+12)/60-(20t^2+40)/60+60/60)#

#I=(sqrt(t^2+2)(3t^4-8t^2+32))/15+C#


However, note that a trig substitution was not actually necessary.

#I=intt^5/sqrt(t^2+2)dt#

Let #u=t^2+2# so #du=2tdt#. This also implies that #t^4=(u-2)^2#.

#I=1/2intt^4/sqrt(t^2+2)(2tdt)=1/2int(u-2)^2/sqrtudu#

Expanding the numerator and dividing each term, then integrating term by term will give the same result with less hassle.

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