# How do you integrate #int 1/sqrt(x^2-a^2)# by trigonometric substitution?

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#### Explanation

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#### Explanation:

#intdx/sqrt(x^2-a^2)#

We will use the substitution

#=int(asecthetatanthetad theta)/sqrt(a^2sec^2theta-a^2)=int(asecthetatanthetad theta)/(asqrt(sec^2theta-1))#

Note that

#=int(secthetatanthetad theta)/tantheta=intsecthetad theta=lnabs(sectheta+tantheta)+C#

From

So,

#=lnabs(x/a+sqrt(x^2-a^2)/a)+C#

Note that a

#=lnabs(x+sqrt(x^2-a^2))+C#

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