Trigonometric Integrals?

Question

Could someone please help out with this question?

fINT(1/[(t^3)(sqrt( t^2 - 1)], x, 2, sqrt(2))

We're supposed to use trigonometric substitutions.

I have literally tried solving this problem for 3 hours straight. No one on my floor can solve it.

Thanks for any help!

Answer 1

I don't get the real question, how can the unknown be x? Zzz the presentation of question is also a bit messy >< (no offense)

is it

∫ 1/[t³ √(t²-1)] dt? then the variable being t? upper limit: 2, lower limit: √2  ?

since it is substitution, did u miss out anything in ur qn? pls

add more details ty (:

lol this is hard...=='''

i'll be trying and thinking as i continue with my math assignment 1st (:

ahh alex got it first (:

i have to take my hats off some for people in this forum...alex is one of em..ingenious gr8 job bro.

anw hw did u see it was substitution of secx? i noe there are only sinx cosx tanx secx cosecx cotx but hw did u draw e relation tt it was secx? pls explain if you can ty (:

Answer 2

Integral of 1 / ( (t^3) (Sqrt (t^2 - 1))) dt from Sqrt (2) to 2.

To easily do this while typing, I will leave off the bounds and the integral sign. I will bring the bounds back later.

1 / ( (t^3) (Sqrt (t^2 - 1))) dt

For trigonometric substitution, you want the square root to go away and then, hopefully it will simplify down to something nice and easy. Note that 1 + Tan(x)^2 = Sec(x)^2, which means that Tan(x)^2 = Sec(x)^2 - 1. If you see this, Sec(x)^2 - 1 looks close to t^2 - 1, so we want to make it look like that.

let sec(x) = t

Taking the derivative

sec(x) tan(x) dx = dt

Now, it is time to plug in the values back into the expression.

1 / (Sec(x)^3 * Sqrt (Sec(x)^2 - 1)) * Sec(x) Tan(x) dx

1 / (Sec(x)^3 * Sqrt (Tan(x)^2)) * Sec(x) Tan(x) dx

1 / (Sec(x)^3 * Tan(x)) * Sec(x) Tan(x) dx

1 / (Sec(x)^2) dx

Cos(x)^2 dx

Also the bounds need to be changed, since it is from t = sqrt(2) to t = 2, plug those values in for t. sec (x) = sqrt(2) ==> x = pi / 4, sec (x) = 2 ==> x = pi / 3. So now the bounds are pi / 4 and pi / 3.

Now, it is a simple matter of taking the integral of Cos(x)^2 and plugging in the newly found limits. I wont go into detail on finding the integral of Cos(x)^2 (it is fairly simple, you just split cos(x)^2 up into (1/2) (cos(2x) + 1) then integrate that. The answer is (1/2) (Sin(x) Cos(x) + x)

(1/2) (Sin(x) Cos(x) + x)

Now all that is left is to plug in the bounds

(1/2) (Sin (pi / 3) Cos(pi / 3) + pi / 3) - (1/2) (Sin(pi / 4) Cos(pi / 4) + pi / 4)

(1/2) (Sin(pi / 3) Cos(pi / 3) + pi / 3 - Sin(pi / 4) Cos (pi / 4) - pi / 4)

Since:

Sin(pi / 3) = Sqrt(3) / 2

Cos(pi / 3) = 1 / 2

Sin(pi / 4) = Sqrt(2) / 2

Cos(pi / 4) = Sqrt(2) / 2

(1/2) ((Sqrt(3) / 2) (1 / 2) + pi / 3 - (Sqrt(2) / 2) (Sqrt(2) / 2) - pi / 4)

(1/2) ((Sqrt(3) / 4) - (2 / 4) - pi / 4+ pi / 3)

(1/2) ((Sqrt(3) / 4) - 1 / 2 + pi / 12)

(after a common denominator)

(3 Sqrt(3) + pi - 6) / 24

Which is approximately equal to 0.097406

Theres how to find the solution using trigonometric substitution.