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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: recovered on June 01, 2011, 05:15:38 pm

Title: anti diff question
Post by: recovered on June 01, 2011, 05:15:38 pm: hey guys so i just want to knw how would u ANTI DIFF this

sqr rt of 4-x^2 dx
Title: Re: anti diff question
Post by: b^3 on June 01, 2011, 05:47:05 pm: 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)
Title: Re: anti diff question
Post by: moekamo on June 01, 2011, 05:48:18 pm: let $x=2\sin(\theta) \implies dx= 2\cos(\theta) d\theta$

so $I=\int \sqrt{4-x^2} dx = \int \sqrt{4-4\sin^2\theta} 2cos(\theta) d\theta = \int 2\cos\theta 2\cos \theta d\theta = \int 4 \cos ^2 \theta d\theta$

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!
Title: Re: anti diff question
Post by: recovered on June 01, 2011, 06:23:28 pm: thank you guys