How do you integrate #int 1/(usqrt(5-u^2))# by trigonometric substitution?

1 Answer
Jun 15, 2017

The answer is #-1/sqrt(5) ln|(sqrt(5) + sqrt(5 - u^2))/u| + C#

Explanation:

Use the substitution #u = sqrt(5)sintheta#. Then #du = sqrt(5)costhetad theta#. Calling the integral #I#, we have:

#I = int 1/(sqrt(5)sin thetasqrt(5 - (sqrt(5)sin theta)^2)) * sqrt(5)costheta d theta#

#I = int 1/(sqrt(5)sin theta sqrt(5 - 5sin^2theta)) * sqrt(5)costheta d theta#

We can now simplify the square root in the denominator.

Since: #sqrt(5 - 5sin^2theta) = sqrt(5(1 - sin^2theta)) = sqrt(5(cos^2theta)) = sqrt(5)costheta# because #sin^2theta + cos^2theta= 1#.

#I = int 1/(sqrt(5)sin theta sqrt(5)costheta) * sqrt(5)costheta d theta#

#I = int1/(sqrt(5)sin theta) d theta#

#I = int1/sqrt(5) csc theta d theta#

The integral of cosecant is known.

#color(blue)(int cscx dx= -ln|cscx + cotx|)#

Therefore:

#I = -1/sqrt(5)ln|csctheta + cottheta|#

From our initial substitution, we know that #sintheta= u/sqrt(5)# and therefore #csctheta = 1/sintheta = sqrt(5)/u#. Also, #cottheta =sqrt(5 - u^2)/u#.

#I = -1/sqrt(5)ln|sqrt(5)/u + sqrt(5 - u^2)/u|#

Simplify, and don't forget to add the constant of integration at the end!

#I = -1/sqrt(5)ln|(sqrt(5) + sqrt(5 - u^2))/u| + C#

Hopefully this helps!