find the integral of {1/sqrt(49-4t^2)} dt?

Answer 1

One standard integral that is best to memorize is that

∫1/sqrt(a^2 - u^2) du = arcsin(u/a) + C,

where a is a positive constant.

In this case let us let

u = 2t
du/dt = 2
dt = du/2
a = 7

So now, our integral becomes

1/2 * ∫ 1/sqrt(a^2-u^2) du
=
1/2 * arcsin(u/a) + C
=
1/2 * arcsin(2t/7) + C.

Answer 2

Use trig substitution: 2t/7 = sin A, or t = (7/2)*sin A and
(sqrt(47 - 4t^2) )/7 = cos A. Therefore, dt = (7/2)*cos A dA. The integrand becomes
(1/2)*dA. The integral is (1/2)*A + C =
(1/2)*arccos(sqrt(49 - 4t^2)/7) + C.

Answer 3

rewrite the expression under the square root as follows:
49 - 4t^2=4(49/4-t^2) .............(1)
let t=(7/2)cos u .............(2)
differentiating both sides of(2),we get:
dt = -(7/2)sin u du ...............(3)
using(2)and(3) in(1),we get:
49 -4t^2=4(49/4 - (49/4) cos^2 u)
=49(1-cos^2 u)
= 49 sin^2 u ...............(4)
using(2) and(4) in the given integrand
integral of {1/sqrt(49-4t^2)} dt =
integral of{1/sqrt(49sin^2 u)}*-(7/2)sin u du
=(-1/2)*integral of{du}
=(-1/2)u +constant ................(5)
from(2) : cos u = 2t/7 ----->
u = arccos(2t/7) ................(6)
using(6) in(5)
the required integral=(-1/2)arccos(2t/7)+ constant.
NOTE: this integral equals (1/2)arcsin(2t/7)+constant.