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Evaluate the integral and express your answer in simplest form. int(-3)/(sqrt(4-16x^(2)))dx Answer:

Evaluate the integral and express your answer in simplest form.\newline∫−34−16x2dx \int \frac{-3}{\sqrt{4-16 x^{2}}} d x \newlineAnswer:

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Q. Evaluate the integral and express your answer in simplest form.\newline∫−34−16x2dx \int \frac{-3}{\sqrt{4-16 x^{2}}} d x \newlineAnswer:
  1. Given Integral: We are given the integral: \newline∫−34−16x2dx\int \frac{-3}{\sqrt{4-16x^{2}}}dx\newlineTo solve this integral, we can use a trigonometric substitution. Let's choose x=12sin⁡(θ)x = \frac{1}{2}\sin(\theta), because the expression 4−16x24 - 16x^2 resembles a Pythagorean identity when xx is expressed in terms of sine.
  2. Trigonometric Substitution: First, we need to find dxdx in terms of dθd\theta. Differentiating x=12sin⁡(θ)x = \frac{1}{2}\sin(\theta) with respect to θ\theta, we get:\newlinedxdθ=12cos⁡(θ)\frac{dx}{d\theta} = \frac{1}{2}\cos(\theta)\newlineTherefore, dx=12cos⁡(θ)dθdx = \frac{1}{2}\cos(\theta)d\theta
  3. Finding dx: Now, we substitute x=12sin⁡(θ)x = \frac{1}{2}\sin(\theta) and dx=12cos⁡(θ)d(θ)dx = \frac{1}{2}\cos(\theta)d(\theta) into the integral:\newline∫−34−16x2dx=∫−34−16(12sin⁡(θ))2(12cos⁡(θ))d(θ)\int\frac{-3}{\sqrt{4-16x^{2}}}dx = \int\frac{-3}{\sqrt{4-16\left(\frac{1}{2}\sin(\theta)\right)^2}}\left(\frac{1}{2}\cos(\theta)\right)d(\theta)\newlineSimplify the expression inside the square root:\newline=∫−34−4sin⁡2(θ)(12cos⁡(θ))d(θ)= \int\frac{-3}{\sqrt{4-4\sin^2(\theta)}}\left(\frac{1}{2}\cos(\theta)\right)d(\theta)\newline=∫−34(1−sin⁡2(θ))(12cos⁡(θ))d(θ)= \int\frac{-3}{\sqrt{4\left(1-\sin^2(\theta)\right)}}\left(\frac{1}{2}\cos(\theta)\right)d(\theta)
  4. Substitute xx and dxdx: Recognize that 1−sin⁡2(θ)1 - \sin^2(\theta) is equal to cos⁡2(θ)\cos^2(\theta) due to the Pythagorean identity. So we can further simplify the integral:\newline=∫−34cos⁡2(θ)(12)cos⁡(θ)d(θ)= \int \frac{-3}{\sqrt{4\cos^2(\theta)}}\left(\frac{1}{2}\right)\cos(\theta)d(\theta)\newline=∫−32∣cos⁡(θ)∣(12)cos⁡(θ)d(θ)= \int \frac{-3}{2|\cos(\theta)|}\left(\frac{1}{2}\right)\cos(\theta)d(\theta)\newlineSince θ\theta is the arcsine of xx, and we are dealing with a square root, we can assume that cos⁡(θ)\cos(\theta) is positive in the range we are considering.\newline=∫−32cos⁡(θ)(12)cos⁡(θ)d(θ)= \int \frac{-3}{2\cos(\theta)}\left(\frac{1}{2}\right)\cos(\theta)d(\theta)\newline=∫(−34)d(θ)= \int \left(-\frac{3}{4}\right)d(\theta)
  5. Simplify Expression: Now we can integrate with respect to θ\theta: ∫(−34)d(θ)=(−34)θ+C\int(-\frac{3}{4})d(\theta) = (-\frac{3}{4})\theta + C
  6. Integrate with respect to theta: We need to express θ\theta back in terms of xx. Since x=12sin⁡(θ)x = \frac{1}{2}\sin(\theta), we have θ=arcsin⁡(2x)\theta = \arcsin(2x). So we substitute back: −34θ+C=−34arcsin⁡(2x)+C-\frac{3}{4}\theta + C = -\frac{3}{4}\arcsin(2x) + C
  7. Express θ\theta in terms of xx: We have found the indefinite integral in terms of xx: ∫−34−16x2dx=(−34)arcsin⁡(2x)+C\int \frac{-3}{\sqrt{4-16x^{2}}}dx = \left(-\frac{3}{4}\right)\arcsin(2x) + C

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