anti diff question

[recovered]

hey guys so i just want to knw how would u ANTI DIFF this

sqr rt of 4-x^2 dx

[b^3]

so sub x=2sin(u) so dx/du=2cos(u)
so dx=2cos(u)*du
so (4-x^2)^(1/2) = (4-4sin^2(u))^(1/2)
=(4(1-sin^2(u))^(1/2)
=2cos(u)
so now  ∫2cos(u)*2cos(u)du since dx=2cos(u)*du
= 4∫cos^2(u)du
= 4∫1/2 + cos(2u)/2du
= 4 ∫1/2du + 4*1/2∫cos(2u)du
= 2u + 2∫cos(2u)du
= 2u + 2*(1/2)sin(2u) +c
= 2u + sin(2u) +c
= 2u + 2sin(u)*cos(u) + c
now sub back in u=sin^-1(x/2)
so 2sin^-1(x/2) + 2sin(sin^-1(x/2))*cos(sin^-1(x/2)
=2sin^-1(x/2) + (x/2)*(4-x^2)^(1/2)
sorry i made a mistake, fixed in the edit
=2sin^-1(x/2) + (1/2)*(4-x^2)^(1/2)

[moekamo]

let

so

should be able to do the rest using spec methods, remember to sub back in at the end to get in terms of x.

ALSO, this is note part of the spec course, i repeat, TRIG substitution like this is not required!

[recovered]

thank you guys