Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 25

To be able to solve for definite integral, we follow the first fundamental theorem of calculus: int_a^b f(x) dx = F(x) +C
such that f is continuous and F is the antiderivative of f in a closed interval [a,b] .
The [a.b] is the boundary limits of the integral such as lower bound=a and upper bound = b.
For the given problem: int_3^(6) 1/(25+(x-3)^2)dx ,
it resembles the basic integration formula:
int (du)/(a^2+u^2) =(1/a)arctan(u/a)+C .
By comparison: (du)/(a^2+u^2) vs(1/(25+(x-3)^2))dx , we may apply
u-substitution by letting:
u^2=(x-3)^2 then u = x-3
where a^2=25 or 5^2 then a=5
Derivative of u will be du = 1 dx or du = dx .
int_3^(6) 1/(25+(x-3)^2)dx =int_3^(6) 1/(25+(u)^2)du
Applying the formula:
int_3^(6) 1/(25+(u)^2)du =(1/5)arctan(u/5)|_3^6
Plug-in u = x-3 to express the indefinite integral in terms of x:
(1/5)arctan(u/5)|_3^6 =(1/5)arctan((x-3)/5)|_3^6
Recall F(x)|_a^b = F(b) - F(a) then:
(1/5)arctan((x-3)/5)|_3^6 = F(6)-F(3)
= (1/5)arctan((6-3)/5) -(1/5)arctan((3-3)/5)
= (1/5)arctan(3/5) -(1/5)arctan(0/5)
=(1/5)arctan(3/5) -0
=(1/5)arctan(3/5) as the Final Answer.

Note: arctan(0/5) = arctan(0)= 0
since tan(theta) = 0 when theta=0

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