How do you integrate #int 1/sqrt(25-t^2)# by trigonometric substitution?

1 Answer
Andrea S.
Mar 19, 2017

#int (dt)/sqrt(25-t^2) = arcsin(t/5) +C#

Explanation:

Substitute:

#t=5sinx#
#dt = 5cosx#

so that:

#int (dt)/sqrt(25-t^2) = int (5cosxdx)/sqrt(25-25sin^2x)#

#int (dt)/sqrt(25-t^2) = int (5cosxdx)/sqrt(25(1-sin^2x))#

#int (dt)/sqrt(25-t^2) = int (cancel5cosxdx)/(cancel5sqrt(cos^2x))#

Now, the integrand function is defined only for #t in (-5,5)#, so that we can assume #x in (-pi/2,pi/2)#. In this interval #cosx# is always positive, so:

#sqrt(cos^2x) = cosx#

#int (dt)/sqrt(25-t^2) = int (cosxdx)/(cosx) = int dx = x+C#

Undoing the substitution:

#int (dt)/sqrt(25-t^2) = arcsin(t/5) +C#

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