Math Problem Statement

Integral(sqrt(x^(2)-1))

Solution

To solve the integral x21dx\int \sqrt{x^2 - 1} \, dx, we can use a trigonometric substitution. Here’s a step-by-step solution:

  1. Trigonometric Substitution: Let x=sec(θ)x = \sec(\theta). Then dx=sec(θ)tan(θ)dθdx = \sec(\theta)\tan(\theta) \, d\theta.

    Also, we have: x21=sec2(θ)1=tan2(θ)=tan(θ)\sqrt{x^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \sqrt{\tan^2(\theta)} = |\tan(\theta)|

    Since θ\theta ranges such that sec(θ)1\sec(\theta) \geq 1, tan(θ)\tan(\theta) is positive, and tan(θ)=tan(θ)|\tan(\theta)| = \tan(\theta).

  2. Substitute and Simplify: Substitute x=sec(θ)x = \sec(\theta) and dx=sec(θ)tan(θ)dθdx = \sec(\theta)\tan(\theta) \, d\theta: x21dx=tan(θ)sec(θ)tan(θ)dθ=sec(θ)tan2(θ)dθ\int \sqrt{x^2 - 1} \, dx = \int \tan(\theta) \cdot \sec(\theta)\tan(\theta) \, d\theta = \int \sec(\theta)\tan^2(\theta) \, d\theta

  3. Simplify the Integral: Use the identity tan2(θ)=sec2(θ)1\tan^2(\theta) = \sec^2(\theta) - 1: sec(θ)tan2(θ)dθ=sec(θ)(sec2(θ)1)dθ\int \sec(\theta)\tan^2(\theta) \, d\theta = \int \sec(\theta)(\sec^2(\theta) - 1) \, d\theta =sec3(θ)dθsec(θ)dθ= \int \sec^3(\theta) \, d\theta - \int \sec(\theta) \, d\theta

  4. Integrate Separately:

    • The integral sec(θ)dθ=lnsec(θ)+tan(θ)+C\int \sec(\theta) \, d\theta = \ln|\sec(\theta) + \tan(\theta)| + C.
    • The integral sec3(θ)dθ\int \sec^3(\theta) \, d\theta can be approached using integration by parts or other techniques, but for now, let's use a known result: sec3(θ)dθ=12sec(θ)tan(θ)+12lnsec(θ)+tan(θ)+C\int \sec^3(\theta) \, d\theta = \frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)| + C
  5. Combine Results: Combine the integrals: sec3(θ)dθsec(θ)dθ=(12sec(θ)tan(θ)+12lnsec(θ)+tan(θ))lnsec(θ)+tan(θ)+C\int \sec^3(\theta) \, d\theta - \int \sec(\theta) \, d\theta = \left(\frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)|\right) - \ln|\sec(\theta) + \tan(\theta)| + C =12sec(θ)tan(θ)12lnsec(θ)+tan(θ)+C= \frac{1}{2} \sec(\theta) \tan(\theta) - \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)| + C

  6. Substitute Back: Substitute back x=sec(θ)x = \sec(\theta) and tan(θ)=x21\tan(\theta) = \sqrt{x^2 - 1}: =12xx2112lnx+x21+C= \frac{1}{2} x \sqrt{x^2 - 1} - \frac{1}{2} \ln|x + \sqrt{x^2 - 1}| + C

Therefore, the integral x21dx\int \sqrt{x^2 - 1} \, dx is: x21dx=12xx2112lnx+x21+C\int \sqrt{x^2 - 1} \, dx = \frac{1}{2} x \sqrt{x^2 - 1} - \frac{1}{2} \ln|x + \sqrt{x^2 - 1}| + C

Would you like more details or have any questions?

Here are five more questions you might be interested in:

  1. How do you solve sin2(x)dx\int \sin^2(x) \, dx?
  2. What is the derivative of ln(x2+1)\ln(x^2 + 1)?
  3. Can you explain the method of partial fractions?
  4. How do you find the limit limx0sin(x)x\lim_{{x \to 0}} \frac{\sin(x)}{x}?
  5. How do you solve a second-order linear differential equation?

Tip: When solving integrals, always check if a trigonometric substitution or another method like integration by parts can simplify the problem.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Integral Calculus
Trigonometric Substitution

Formulas

Trigonometric substitution formulas

Theorems

-

Suitable Grade Level

Advanced College Level