How do you find:
the Integral of arcsin(1 / (sqrt x^2 - 1) ) dx ??
(The integral of arcsin of one over the squareroot of x squared minus 1)
So far, I've used trig integration to simplify down to
arcsin(1/tanb) *secbtanb
b is theta
Then, I've used integration by parts. But i'm getting very stuck.
Please help? I've been working on this problem FOREVER.
Thank you!
To find the integral of arcsin(1 / (sqrt(x^2 - 1))), you made a good first step by using trigonometric identities to simplify the expression. Let's go through the process step by step.
Start with the original integral:
∫arcsin(1 / √(x^2 - 1)) dx
First, we'll make a substitution to simplify the expression. Let's assume:
x = sec(b)
Differentiating both sides with respect to b using the chain rule, we get:
dx = sec(b)tan(b) db
Now we can rewrite the integral in terms of b:
∫arcsin(1 / √(sec^2(b) - 1)) sec(b)tan(b) db
Let's work on simplifying the expression inside the arcsin function. Recall the trig identity: sec^2(b) - 1 = tan^2(b). Replacing sec^2(b) - 1 with tan^2(b), we have:
∫arcsin(1 / √(tan^2(b))) sec(b)tan(b) db
Simplifying further:
∫arcsin(1 / tan(b)) sec(b)tan(b) db
Now, we have an integral with arcsin(1 / tan(b)), sec(b), and tan(b). We can use integration by parts to solve this integral.
Recall the integration by parts formula:
∫u dv = uv - ∫v du
Let's assign u and dv:
u = arcsin(1 / tan(b))
dv = sec(b)tan(b) db
Taking the derivatives and integrals:
du = (1 / √(1 - (1 / tan^2(b)))) (tan(b))^-2 db
= (cos(b) / √(1 - 1 / sin^2(b))) (tan(b))^-2 db
= cos(b) / √((sin^2(b) - 1) / sin^2(b)) (tan(b))^-2 db
= cos(b) / √((-cos^2(b)) / sin^2(b)) (tan(b))^-2 db
= sin(b) / √(cos^2(b)) (tan(b))^-2 db
= sin(b) / (cos(b) * tan^2(b)) db
= sin(b) / (cos(b) / sin^3(b)) db
= sin^4(b) / cos(b) db
v = ∫sec(b)tan(b) db
= ∫sec(b) * sin(b)/cos(b) db
= ∫tan(b) db
= -ln|cos(b)|
Now we can evaluate the integral using the integration by parts formula:
∫arcsin(1 / tan(b)) sec(b)tan(b) db = (-ln|cos(b)|) * arcsin(1 / tan(b)) - ∫(-ln|cos(b)|) * (sin^4(b) / cos(b)) db
Now we have a new integral to solve, which is a bit simpler than the original one.
I hope this explanation helps you progress with your integral. If you have any further questions, please feel free to ask!
To find the integral of arcsin(1 / (sqrt(x^2 - 1))) dx, you're on the right track using trigonometric substitution. Let's continue step-by-step.
1. Start by letting x = sec(θ). Then, dx = sec(θ)tan(θ) dθ.
2. Rewrite the integrand in terms of θ:
arcsin(1 / (sqrt((sec^2(θ)) - 1))) = arcsin(1 / (sqrt(tan^2(θ)))).
3. Simplify the integrand:
arcsin(1 / (tan(θ))) = arcsin(cot(θ)) = θ.
4. Now, we need to determine the limits of integration in terms of θ.
When x = 1, we have sec(θ) = 1, which implies θ = 0.
When x approaches infinity, sec(θ) approaches infinity as well. In this case, we have θ = π/2.
5. Finally, rewrite the integral using the new variables:
∫ arcsin(1 / (sqrt(x^2 - 1))) dx = ∫ θ(sec(θ)tan(θ)) dθ.
6. Integrate with respect to θ:
∫ θ(sec(θ)tan(θ)) dθ = ∫ θ(1/cos(θ))(sin(θ)/cos(θ)) dθ.
7. Simplify the integrand:
∫ θ(sin(θ) / cos^2(θ)) dθ.
8. Integrate using a u-substitution, where u = cos(θ):
∫ (tan(u) / u^2) du.
This integral can be evaluated using various techniques, such as integration by parts or using a Taylor series expansion. Unfortunately, it does not have a simple closed-form solution.
So, the final result is the indefinite integral of arcsin(1 / (sqrt(x^2 - 1))) dx is expressed as ∫ θ(sec(θ)tan(θ)) dθ, where the evaluation of this integral depends on how you choose to handle the final integral (such as using numerical methods or approximations).
I hope this helps! Let me know if you have any further questions.