do the following 3 integrals?

Question

integral of sqrt(1-x^2)dx
integral of sqrt(x^2-1)dx
integral of sqrt(x^2+1)dx

no formulas...need to prove them :( PLease help its 3:45 am and I've been up all night!

yes that helps but since my brain is dead :P i need more help. There are no 1's on top of the integrals and thats messing me up. There is a nasty formula in the back of the book as the answer, but I have to get from these integrals to that answer by myself...and i cant boast about being a calc genius...thats wht i need some help :(

Answer 1

go to bed; come back in 2 hours! they will be done;

♣ thus y(x)*dx = dx*√(1-x^2);
x=sin(t), dx=cos(t)*dt, t=asin(x);
y(t)*dt = cos(t)*dt *√(1-(sin(t))^2) =
= dt*(cos t)^2 = dt *0.5*(cos(2t) +1) =
= 0.5*dt * cos(2t) + 0.5*dt;
Y(t) = 0.25*sin(2t) +0.5*t =
= 0.5 * sin t *cos t +0.5*t =
= 0.5*x*√(1-x^2) +0.5* asin(x) +C;
♣ thus y(x)*dx = dx*√(x^2 -1);
x= ch(t), dx=dt*sh(t),
t= -ln(x -√(x^2 -1)) =ln(x+√(x^2 -1));
y(t)*dt = dt*sh(t) *√((ch t)^2 -1) =
= dt*(sh(t))^2 = 0.5*dt*(ch(2t) -1);
Y(t) = 0.25*sh(2t) -0.5*t =
= 0.5 * sh t *ch t -0.5*t =
= 0.5*x*√(x^2 -1) +0.5*ln(x -√(x^2 -1)) +C;
♣ thus y(x)*dx = dx*√(x^2 +1);
x= sh(t), dx=dt*ch(t), t= ln(x +√(x^2 +1));
y(t)*dt = dt*ch(t) *√((sh t)^2 +1) =
= dt*(ch(t))^2 = 0.5*dt*(ch(2t) +1);
Y(t) = 0.25*sh(2t) +0.5*t =
= 0.5 * sh t *ch t +0.5*t =
= 0.5*x*√(x^2 +1) +0.5*ln(x +√(x^2 +1)) +C;

Answer 2

The following are listed in text books as standard integrals:-

∫ 1 / √(1 - x ²) dx = sin^(-1) x + C

∫ 1 / √ (x ² - 1) dx = cosh ^(-1) x + C

∫1 / √(x² + 1) dx = sinh^(-1) x + C

Does this help?

Alternatively consider
(1 - x²)^(1/2) and expand using something like Taylor series.
Then integrate each term in the series to obtain required answer.