integrate:dx/((x-1)sqrt(x^2-2)
plz show solution i plead
see the steps here.
http://www.integral-calculator.com/
Interestingly, wolframalpha.com simplifies it a lot.
Sir it won,t solve it for me i tried it....i don,t even no how to use it...plz help me
dx/((x-1)sqrt(x^2-2))
OK. First step is a trig substitution.
x = √2 sec(u)
x^2-2 = 2tan^2(u)
dx = √2 sec(u) tan(u) du
now the integral is
sec(u)/(√2 sec(u)-1) du
see what you can do with that.
To integrate the given expression:
∫ dx/((x-1)√(x^2-2))
Let's start by simplifying the expression inside the square root:
√(x^2-2) = √((x√2)^2-2)
= √(x^2√2^2-2)
= √(2x^2-2)
Now let's rewrite the integral:
∫ dx/((x-1)√(2x^2-2))
Next, we can apply a trigonometric substitution. Let's try substituting x = √2secθ:
dx = √2secθtanθ dθ
√(2x^2-2) = √(2(√2secθ)^2-2)
= √(2(2sec^2θ)-2)
= √(4sec^2θ-2)
= √(4(tan^2θ+1)-2)
= √(4tan^2θ+2)
Now let's re-write the integral using the substitution:
∫ √2secθtanθ dθ / ((√2secθ)-1)√(4tan^2θ+2)
Simplifying, we get:
∫ √2secθtanθ dθ / (√2secθ-1) √(4tan^2θ+2)
= ∫ √2tanθ dθ / (√2secθ-1) √(4tan^2θ+2)
Now, we need to express everything in terms of θ. To do this, we can use the identity: secθ = 1/cosθ.
Let's re-write the expression using this identity:
∫ √2tanθ dθ / (√2/cosθ-1) √(4tan^2θ+2)
= ∫ √2tanθ dθ ∙ (√2cosθ - 1) / (√2√(4tan^2θ+2))
Simplifying further:
= ∫ √2√2tanθcosθ - √2tanθ dθ / √(8tan^2θ+4)
Now, we can simplify the expression:
= ∫ 2tanθcosθ dθ - ∫ √2tanθ dθ / √(8tan^2θ+4)
The integral of 2tanθcosθ can be solved using the substitution method or by using the trigonometric identity tanθ = sinθ/cosθ.
After solving both integrals, we can substitute the original variable x back in for θ.
Please note that the solution may be quite involved and may require additional use of trigonometric identities and integration techniques.