Evaluate the integral: int_(0)^(2)(dx)/(sqrt(x^(2)+4))

Trig Substitution: We are given the integral to evaluate: int_(0)^(2)(dx)/(sqrt(x^(2)+4)) To solve this integral, we can use a trigonometric substitution. Let's choose x = 2tan(θ), where θ is the new variable of integration. This substitution is chosen because it will simplify the square root in the denominator.Find dx in terms of dθ: First, we need to find the differential dx in terms of dθ. Since x = 2tan(θ), we take the derivative with respect to θ to get: dx/dθ = 2sec^2(θ) Therefore, dx = 2sec^2(θ)dθSubstitute x = 2tan(θ): Now we substitute x = 2tan(θ) into the integral and change the limits of integration accordingly. When x = 0, tan(θ) = 0, so θ = 0. When x = 2, tan(θ) = 1, so θ = π/4. The integral becomes: int_(0)^(π/4)(2sec^2(θ)dθ)/(sqrt((2tan(θ))^2+4))Simplify the integral: Simplify the integral by substituting x = 2tan(θ) and dx = 2sec^2(θ)dθ: int_(0)^(π/4)(2sec^2(θ)dθ)/(sqrt(4tan^2(θ)+4)) = int_(0)^(π/4)(2sec^2(θ)dθ)/(sqrt(4(tan^2(θ)+1))) = int_(0)^(π/4)(2sec^2(θ)dθ)/(2sqrt(tan^2(θ)+1))Use trigonometric identities: We know that sec^2(θ) - tan^2(θ) = 1, so sqrt(tan^2(θ) + 1) = sec(θ). The integral simplifies to: int_(0)^(π/4)(2sec^2(θ)dθ)/(2sec(θ)) = int_(0)^(π/4)sec(θ)dθEvaluate the integral: The integral of sec(θ) is ln|sec(θ) + tan(θ)|. So we evaluate the integral from 0 to π/4: int_(0)^(π/4)sec(θ)dθ = ln|sec(π/4) + tan(π/4)| - ln|sec(0) + tan(0)|Plug in values: Now we plug in the values for θ = π/4 and θ = 0: ln|sec(π/4) + tan(π/4)| - ln|sec(0) + tan(0)| = ln|√2 + 1| - ln|1 + 0| = ln(√2 + 1) - ln(1) = ln(√2 + 1)

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Evaluate the integral: int_(0)^(2)(dx)/(sqrt(x^(2)+4))

Evaluate the integral: $\int_{0}^{2} \frac{d x}{\sqrt{x^{2}+4}}$

View step-by-step help

Home

Math Problems

Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Evaluate the integral:

\int_{0}^{2} \frac{d x}{\sqrt{x^{2}+4}}

Trig Substitution: We are given the integral to evaluate: $\newline$ $\int_{0}^{2}\frac{dx}{\sqrt{x^{2}+4}}$ $\newline$ To solve this integral, we can use a trigonometric substitution. Let's choose $x = 2\tan(\theta)$ , where $\theta$ is the new variable of integration. This substitution is chosen because it will simplify the square root in the denominator.
Find $dx$ in terms of $d\theta$ : First, we need to find the differential $dx$ in terms of $d\theta$ . Since $x = 2\tan(\theta)$ , we take the derivative with respect to $\theta$ to get: $\newline$ $\frac{dx}{d\theta} = 2\sec^2(\theta)$ $\newline$ Therefore, $dx = 2\sec^2(\theta)d\theta$
Substitute $x = 2\tan(\theta)$ : Now we substitute $x = 2\tan(\theta)$ into the integral and change the limits of integration accordingly. When $x = 0$ , $\tan(\theta) = 0$ , so $\theta = 0$ . When $x = 2$ , $\tan(\theta) = 1$ , so $\theta = \frac{\pi}{4}$ . The integral becomes: $\int_{0}^{\frac{\pi}{4}}\frac{2\sec^2(\theta)d\theta}{\sqrt{(2\tan(\theta))^2+4}}$
Simplify the integral: Simplify the integral by substituting $x = 2\tan(\theta)$ and $dx = 2\sec^2(\theta)d\theta$ : $\int_{0}^{\frac{\pi}{4}}\frac{2\sec^2(\theta)d\theta}{\sqrt{4\tan^2(\theta)+4}}$ = $\int_{0}^{\frac{\pi}{4}}\frac{2\sec^2(\theta)d\theta}{\sqrt{4(\tan^2(\theta)+1)}}$ = $\int_{0}^{\frac{\pi}{4}}\frac{2\sec^2(\theta)d\theta}{2\sqrt{\tan^2(\theta)+1}}$
Use trigonometric identities: We know that $\sec^2(\theta) - \tan^2(\theta) = 1$ , so $\sqrt{\tan^2(\theta) + 1} = \sec(\theta)$ . The integral simplifies to: $\newline$ \int_{ $0$ }^{\frac{\pi}{ $4$ }}\frac{ $2$ \sec^ $2$ (\theta)d\theta}{ $2$ \sec(\theta)} $\newline$ = \int_{ $0$ }^{\frac{\pi}{ $4$ }}\sec(\theta)d\theta
Evaluate the integral: The integral of $\sec(\theta)$ is $\ln|\sec(\theta) + \tan(\theta)|$ . So we evaluate the integral from $0$ to $\frac{\pi}{4}$ : $\int_{0}^{\frac{\pi}{4}}\sec(\theta)d\theta = \ln|\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| - \ln|\sec(0) + \tan(0)|$
Plug in values: Now we plug in the values for $\theta = \frac{\pi}{4}$ and $\theta = 0$ : $\newline$ $\ln|\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| - \ln|\sec(0) + \tan(0)|$ $\newline$ $= \ln|\sqrt{2} + 1| - \ln|1 + 0|$ $\newline$ $= \ln(\sqrt{2} + 1) - \ln(1)$ $\newline$ $= \ln(\sqrt{2} + 1)$