Trigonometry substituting (1 Viewer)

[sneaky pete]

hi
sorry im not sure which maths this is, 3unit, 4unit, or extra-curricular, so sorry about that

anyway we have


/ x^2 .dx
| sqrt(4-x^2)
/

allow x = 2sinu
dx = 2cosu du

/ 4sin^2(u) .2cosu du
| 2cos u
/

1) is the reason we cant just cancel off thos 2cos(u)'s is because at the numerator, the 2cos(u) is kind of 'connected' to the du?

2) further down in the solution, it has:
(1) 2u - 2sin(u) cos(u) + C
(2) = 2sin^-1(x/2) - 1/2 x sqrt(4-x^2) + C

-where did that inverse sin come from? and just basicly how did they get from line (1) to (2) ?

thanks

 

[jayz]

u = arcsin(x/2) (arcsin is inverse sin)


cos(u) can be treated separate from du

 

[Heinz]

they do cancel it out the 2cosu and because 4sin^2(u) = 2(1-cos2u), integrating w.r.t.u you get 2u - sin2u = 2u - 2sinucosu
Also, because its an indefinite integral, you gotta resubstitute back in terms of x. so remembering your substitution x = 2sinu :. u = sin^-1(x/2)