How do you integrate #int cost/sqrt(1+sin^2t)# by trigonometric substitution?

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Answer 1 · 2016-09-11T04:12:35

#lnabs(sqrt(sin^2t+1)+sint)+C#

We have:

#intcost/sqrt(1+sin^2t)dt#

We will use the substitution #sint=tantheta#. Thus, #costdt=sec^2thetad theta#.

Substituting, this yields:

#=int(sec^2thetad theta)/sqrt(1+tan^2theta)#

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

#=int(sec^2thetad theta)/sectheta=intsecthetad theta#

This is a common integral:

#=lnabs(sectheta+tantheta)#

Since we know #sint=tantheta#, write #sectheta# in terms of #tantheta#:

#=lnabs(sqrt(tan^2theta+1)+tantheta)#

#=lnabs(sqrt(sin^2t+1)+sint)+C#