Expert: Abe Mantell - 8/21/2010

Question
Dear Expert. I am battling to study due to a rare eye problem. Please can you show me how to solve these integrals. It is new to me and I actually have no idea what is going on. I have 1st year calculus but I am struggling with calculus and extra baggage.
Determine the integrals?
(a) I = ∫√(9 – x^2) dx

(b) I = ∫dx/√(x^2 + 6x +13)

(c) I = ∫ from 1 to 0 x.e^(x^2) dx

Thank you so much

Answer
Hello Peter,

a) ∫√(9 – x^2) dx, make the substitution x=3sin(u)
==> x^2=9sin^2(u) and dx=3cos(u) du, the integral becomes:
.   ∫√[9 – 9sin^2(u)]*3cos(u) du
. =3∫√[9(1 – sin^2(u))]*cos(u) du
. =3∫3√[cos^2(u)]*cos(u) du
. =9∫cos(u)*cos(u) du = 9∫cos^2(u) du
. now use the identity: cos^2(u)=(1/2)[1+cos(2u)]
==> 9∫(1/2)[1+cos(2u)] du = (9/2)∫[1+cos(2u)] du =
.   = (9/2)[u+(1/2)sin(2u)]+C, now use the double angle formula:
.   sin(2u)=2cos(u)sin(u), with sin(u)=x/3
==> u=arcsin(x/3) and cos(u)=sqrt(1-x^2/9)
.   so we get: (9/2)[arcsin(x/3)+(1/2)*2*sqrt(1-x^2/9)*x/3]+C
.   =(9/2)arcsin(x/3)+(1/2)x*sqrt(9-x^2)+C

b) ∫dx/√(x^2 + 6x +13) = ∫dx/√[(x + 3)^2 + 4), now let x+3=2tan(u)
.  so we get dx=2sec^2(u) du and the denominator simplifies
.  sqrt(4(tan^2(u)+1))=sqrt(4)*sqrt(sec^2(u))=2sec(u)
.  we now have ∫2sec^2(u) du/[2sec(u)] = ∫sec(u) du
.  which gives ln|sec(u)+tan(u)| + C
.  = ln((1/2)*x+3/2+(1/2)*sqrt(x^2+6*x+13)) + C

c) make the substitution u=x^2, then I think you can finish it off.
.  You should get 1/2-(1/2)*e...from 1 to 0...if from 0 to 1, then
.  you will get (1/2)e-1

Abe