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Evaluate the integral and express your answer in simplest form. int(-1)/(sqrt(25-9x^(2)))dx Answer:

Evaluate the integral and express your answer in simplest form.\newline∫−125−9x2dx \int \frac{-1}{\sqrt{25-9 x^{2}}} d x \newlineAnswer:

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Q. Evaluate the integral and express your answer in simplest form.\newline∫−125−9x2dx \int \frac{-1}{\sqrt{25-9 x^{2}}} d x \newlineAnswer:
  1. Given Integral: We are given the integral: \newline∫−125−9x2dx\int \frac{-1}{\sqrt{25-9x^{2}}}dx\newlineTo solve this integral, we can use a trigonometric substitution. Let's choose x=53sin⁡(θ)x = \frac{5}{3}\sin(\theta), because the expression 25−9x225 - 9x^2 resembles a Pythagorean identity when xx is expressed in terms of sine.
  2. Trigonometric Substitution: Differentiate x=53sin⁡(θ)x = \frac{5}{3}\sin(\theta) with respect to θ\theta to find dxdx:dxdθ=53cos⁡(θ)\frac{dx}{d\theta} = \frac{5}{3}\cos(\theta)dx=53cos⁡(θ)dθdx = \frac{5}{3}\cos(\theta)d\theta
  3. Differentiate xx: Substitute x=53sin⁡(θ)x = \frac{5}{3}\sin(\theta) and dx=53cos⁡(θ)dθdx = \frac{5}{3}\cos(\theta)d\theta into the integral:\newline∫−125−9x2dx=∫−125−9(53)2sin⁡2(θ)(53)cos⁡(θ)dθ\int \frac{-1}{\sqrt{25-9x^{2}}}dx = \int \frac{-1}{\sqrt{25-9\left(\frac{5}{3}\right)^2\sin^2(\theta)}}\left(\frac{5}{3}\right)\cos(\theta)d\theta\newlineSimplify the expression inside the square root:\newline25−9(53)2sin⁡2(θ)=25−25sin⁡2(θ)=25(1−sin⁡2(θ))25 - 9\left(\frac{5}{3}\right)^2\sin^2(\theta) = 25 - 25\sin^2(\theta) = 25(1 - \sin^2(\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) by the Pythagorean identity:\newline25(1−sin⁡2(θ))=25cos⁡2(θ)25(1 - \sin^2(\theta)) = 25\cos^2(\theta)\newlineNow, substitute this back into the integral:\newline∫(−1)/(25−9∗(5/3)2sin⁡2(θ))∗(5/3)cos⁡(θ)dθ=∫(−1)/(25cos⁡2(θ))∗(5/3)cos⁡(θ)dθ\int(-1)/(\sqrt{25-9*(5/3)^2\sin^2(\theta)})*(5/3)\cos(\theta)d\theta = \int(-1)/(\sqrt{25\cos^2(\theta)})*(5/3)\cos(\theta)d\theta
  5. Simplify Expression: Simplify the integral by taking the square root of 25cos⁡2(θ)25\cos^2(\theta), which is 5cos⁡(θ)5\cos(\theta):\newline∫−15cos⁡(θ)(53cos⁡(θ))dθ\int\frac{-1}{5\cos(\theta)}\left(\frac{5}{3}\cos(\theta)\right)d\theta\newlineThe cos⁡(θ)\cos(\theta) terms cancel out, and we are left with:\newline∫(−13)dθ\int\left(-\frac{1}{3}\right)d\theta
  6. Recognize Identity: Integrate −13-\frac{1}{3} with respect to θ\theta:∫(−13)dθ=−13×θ+C\int\left(-\frac{1}{3}\right)d\theta = -\frac{1}{3} \times \theta + C
  7. Simplify Integral: Now we need to express θ\theta back in terms of xx. Since x=53sin⁡(θ)x = \frac{5}{3}\sin(\theta), we can solve for θ\theta using the inverse sine function:\newlinesin⁡(θ)=35x\sin(\theta) = \frac{3}{5}x\newlineθ=arcsin⁡(35x)\theta = \arcsin(\frac{3}{5}x)\newlineSubstitute θ\theta back into the integral result:\newline−13⋅θ+C=−13⋅arcsin⁡(35x)+C-\frac{1}{3} \cdot \theta + C = -\frac{1}{3} \cdot \arcsin(\frac{3}{5}x) + C

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