y = 2 Sqrt x from o to 1 (Curve , and Area)
- Thread starter Riazy
- Start date
D
Deleted member 4993
Guest
Riazy said:
Substitute
u = sec(?)
Don't forget to change the limits
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,962
If you were to substitute sec(?) for u, then you would have sec(?)^2.
There is a Pythagorean trig-identity that allows us to express sec(?)^2 in terms of the tangent function.
Are you familiar with it?
Last edited:
D
Deleted member 4993
Guest
Well - then I don't know how to do it.
After substitution and back substitution the answer would be
1/2*[x?(x[sup:36k0m72k]2[/sup:36k0m72k]-1)-log{2[?(x[sup:36k0m72k]2[/sup:36k0m72k]-1)+x]}] + C
Strange - scandanavian countries score very high in math competitions.
After substitution and back substitution the answer would be
1/2*[x?(x[sup:36k0m72k]2[/sup:36k0m72k]-1)-log{2[?(x[sup:36k0m72k]2[/sup:36k0m72k]-1)+x]}] + C
Strange - scandanavian countries score very high in math competitions.
For integrals of the form \(\displaystyle \int\sqrt{u^{2}-a^{2}}du\)
use the hyperbolic sub \(\displaystyle u=a\cdot cosh(t), \;\ du=a\cdot sinh(t)dt\)
In your case, a=1.
They come in handy for integrals of this form.
Makes for an easier integration than with the sec sub anyway. At least, I always thought so.
Continuing:
\(\displaystyle \frac{\pi}{2}\int_{1}^{3}\sqrt{u^{2}-1}du\)
Make the sub \(\displaystyle u=cosh(t), \;\ du=sinh(t)dt\)
\(\displaystyle \frac{\pi}{2}\int_{0}^{cosh^{-1}(3)}sinh^{2}(t)dt\)
\(\displaystyle \frac{\pi}{2}\int_{0}^{cosh^{-1}(3)}\frac{1}{2}(cosh(2t)-1)dt\)
\(\displaystyle \frac{\pi}{2}\left(\frac{1}{4}sinh(2t)-\frac{t}{2}\right)\left|_{0}^{cosh^{-1}(3)}\right\)
\(\displaystyle \approx 5.28\)
Do y'all use hyperbolic trig in Sweden?.
use the hyperbolic sub \(\displaystyle u=a\cdot cosh(t), \;\ du=a\cdot sinh(t)dt\)
In your case, a=1.
They come in handy for integrals of this form.
Makes for an easier integration than with the sec sub anyway. At least, I always thought so.
Continuing:
\(\displaystyle \frac{\pi}{2}\int_{1}^{3}\sqrt{u^{2}-1}du\)
Make the sub \(\displaystyle u=cosh(t), \;\ du=sinh(t)dt\)
\(\displaystyle \frac{\pi}{2}\int_{0}^{cosh^{-1}(3)}sinh^{2}(t)dt\)
\(\displaystyle \frac{\pi}{2}\int_{0}^{cosh^{-1}(3)}\frac{1}{2}(cosh(2t)-1)dt\)
\(\displaystyle \frac{\pi}{2}\left(\frac{1}{4}sinh(2t)-\frac{t}{2}\right)\left|_{0}^{cosh^{-1}(3)}\right\)
\(\displaystyle \approx 5.28\)
Do y'all use hyperbolic trig in Sweden?.
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,962
Riazy said:
Then you must be familiar with sine, cosine, and tangent, yes?
Each of those three trigonometric functions has a reciprocal, and the reciprocals are named as the cosecant, secant, and cotangent functions, respectively.
csc = 1/sin
sec = 1/cos
cot = 1/tan
There are three Pythagorean trig identities:
sin^2 + cos^2 = 1
1 + cot^2 = csc^2
tan^2 + 1 = sec^2
Last edited: