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Math  /  Calculus

QuestionEvaluate the definite integral. ∫181x16x2−3dx\int_{1}^{8} \frac{1}{x \sqrt{16 x^{2}-3}} d x

Studdy Solution

STEP 1

1. We are given the definite integral ∫181x16x2−3 dx \int_{1}^{8} \frac{1}{x \sqrt{16 x^{2}-3}} \, dx .
2. We need to evaluate this integral over the interval from x=1 x = 1 to x=8 x = 8 .

STEP 2

1. Simplify the integrand by identifying a suitable substitution.
2. Perform the substitution and change the limits of integration.
3. Integrate the resulting expression.
4. Substitute back to the original variable, if necessary.
5. Evaluate the definite integral using the new limits.

STEP 3

Identify a suitable substitution. Notice the expression under the square root: 16x2−3 16x^2 - 3 . This suggests a trigonometric substitution or a hyperbolic substitution might be useful. Let's try the substitution x=14sec⁡(θ) x = \frac{1}{4} \sec(\theta) .

STEP 4

Differentiate the substitution to find dx dx :
dx=14sec⁡(θ)tan⁡(θ) dθ dx = \frac{1}{4} \sec(\theta) \tan(\theta) \, d\theta
Substitute x=14sec⁡(θ) x = \frac{1}{4} \sec(\theta) into the integrand:
16x2−3=16(14sec⁡(θ))2−3=sec⁡2(θ)−3 \sqrt{16x^2 - 3} = \sqrt{16\left(\frac{1}{4} \sec(\theta)\right)^2 - 3} = \sqrt{\sec^2(\theta) - 3}
Since sec⁡2(θ)−1=tan⁡2(θ) \sec^2(\theta) - 1 = \tan^2(\theta) , we have:
sec⁡2(θ)−3=tan⁡2(θ)−2 \sqrt{\sec^2(\theta) - 3} = \sqrt{\tan^2(\theta) - 2}

STEP 5

Change the limits of integration. When x=1 x = 1 , sec⁡(θ)=4 \sec(\theta) = 4 so θ=sec⁡−1(4) \theta = \sec^{-1}(4) . When x=8 x = 8 , sec⁡(θ)=32 \sec(\theta) = 32 so θ=sec⁡−1(32) \theta = \sec^{-1}(32) .

STEP 6

Substitute into the integral:
∫sec⁡−1(4)sec⁡−1(32)114sec⁡(θ)⋅tan⁡2(θ)−2⋅14sec⁡(θ)tan⁡(θ) dθ \int_{\sec^{-1}(4)}^{\sec^{-1}(32)} \frac{1}{\frac{1}{4} \sec(\theta) \cdot \sqrt{\tan^2(\theta) - 2}} \cdot \frac{1}{4} \sec(\theta) \tan(\theta) \, d\theta
Simplify the expression:
∫sec⁡−1(4)sec⁡−1(32)tan⁡(θ)tan⁡2(θ)−2 dθ \int_{\sec^{-1}(4)}^{\sec^{-1}(32)} \frac{\tan(\theta)}{\sqrt{\tan^2(\theta) - 2}} \, d\theta

STEP 7

Recognize that the integral can be simplified further by using a trigonometric identity or a direct integration technique.
∫tan⁡(θ)tan⁡2(θ)−2 dθ=∫d(tan⁡(θ))tan⁡2(θ)−2 \int \frac{\tan(\theta)}{\sqrt{\tan^2(\theta) - 2}} \, d\theta = \int \frac{d(\tan(\theta))}{\sqrt{\tan^2(\theta) - 2}}
This integral can be solved using a direct substitution or recognizing it as a standard form.

STEP 8

Evaluate the definite integral using the new limits:
[ln⁡∣tan⁡(θ)+tan⁡2(θ)−2∣]sec⁡−1(4)sec⁡−1(32) \left[ \ln \left| \tan(\theta) + \sqrt{\tan^2(\theta) - 2} \right| \right]_{\sec^{-1}(4)}^{\sec^{-1}(32)}
Calculate the values at the limits and subtract:
ln⁡∣tan⁡(sec⁡−1(32))+tan⁡2(sec⁡−1(32))−2∣−ln⁡∣tan⁡(sec⁡−1(4))+tan⁡2(sec⁡−1(4))−2∣ \ln \left| \tan(\sec^{-1}(32)) + \sqrt{\tan^2(\sec^{-1}(32)) - 2} \right| - \ln \left| \tan(\sec^{-1}(4)) + \sqrt{\tan^2(\sec^{-1}(4)) - 2} \right|
The value of the definite integral is:
ln⁡∣tan⁡(sec⁡−1(32))+tan⁡2(sec⁡−1(32))−2tan⁡(sec⁡−1(4))+tan⁡2(sec⁡−1(4))−2∣ \boxed{\ln \left| \frac{\tan(\sec^{-1}(32)) + \sqrt{\tan^2(\sec^{-1}(32)) - 2}}{\tan(\sec^{-1}(4)) + \sqrt{\tan^2(\sec^{-1}(4)) - 2}} \right|}

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