How do you integrate #int 1/(usqrt(5-u^2))# by trigonometric substitution?
1 Answer
The answer is
Explanation:
Use the substitution
#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:
#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
#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!