integrals. so difficult, please help!?

Question

how do i evaluate the following??

integral (0 to 2) of dx/(1+4x^2)

also

integral (0 to 3) of dx/(root(36-x^2))

thanks heaps for the help!

for the question asked below:

it is the integral of (dx) / (1+4x^2)

and (dx) / (root(36-x^2))

Answer 1

some basic things about integrals: an integral is like a reverse derivative. if you take the derivative of a function, you can "undo" this by taking the integral of it. the integral is the measure of the area under a given curve. the fundamental theorem of calculus states that the integral of a fuction f'(x)=F(b)-F(a) , provided the limits of integration are from a to b. the first step is therefore to calculuate the antiderivative of the function. i would help you on the first one, but is the integral 1/(1+4x^2) or is it just (1+4x^2)?

Answer 2

Integral (0 to 2, 1/(1 + 4x^2) dx )

Use trig substitution.
Let x = (1/2)tan(t)
dx = (1/2)sec^2(t) dt

When x = 0, then
(1/2)tan(t) = 0
tan(t) = 0
t = 0

When x = 2, then
(1/2)tan(t) = 2
tan(t) = 4
t = arctan(4)

Integral ( 0 to arctan(4), 1/(1 + tan^2(t)) (1/2)sec^2(t) dt )

Integral ( 0 to arctan(4), 1/sec^2(t) (1/2) sec^2(t) dt )

Integral ( 0 to arctan(4), (1/2) dt )

Which is easy to integrate

(1/2)t {from 0 to arctan(4)}

(1/2)(arctan(4)) - (1/2)(0)

(1/2)arctan(4)

Answer 3

Question 1
I = ∫ dx / 4.[ (1/2)² + x² ]
I = (2/4).tan^(-1) (x /(1/2)) lims 0 to 2
I = (1/2).tan^(-1) (4)
I = 2.65 radians

Question 2
I = ∫ dx / √ [ 6² - x² ]
I = sin^(-1) (x / 6) lims. 0 to 3
I = sin^(-1) (1/2) - 0
I = π / 6 radians

Answer 4

the first answer is 12/2/3 units^2
the second one is 13/1/2 units^2(not sure)