How do you integrate #int 1/(sqrtx(1+x))# by trigonometric substitution?

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Answer 1 · 2016-09-11T04:20:06

#2arctan(sqrtx)+C#

We have the integral:

#intdx/(sqrtx(1+x))#

We will use the substitution #sqrtx=tantheta#. This implies that #x=tan^2theta# and that #1/(2sqrtx)dx=sec^2thetad theta#. Rearranging:

#=2intdx/(2sqrtx(1+x))=2int(sec^2thetad theta)/(1+tan^2theta)#

Through the Pythagorean identity, #1+tan^2theta=sec^2theta#.

#=2intsec^2theta/sec^2thetad theta=2intd theta=2theta+C#

From #sqrtx=tantheta# we see that #theta=arctan(sqrtx)#. Thus:

#=2arctan(sqrtx)+C#