int(dx)/(sqrt(-2x^(2)+8x+4))

Given Integral: We are given the integral to evaluate: \[ \int \frac{dx}{\sqrt{-2x^2 + 8x + 4}} \] First, we need to complete the square for the quadratic expression under the square root to simplify the integral. The quadratic expression is -2x^2 + 8x + 4. To complete the square, we factor out the coefficient of x^2, which is -2: \[ -2(x^2 - 4x - 2) \] Now, we find the number to complete the square, which is $\left(\frac{4}{2}\right)^2 = 4$. We add and subtract this number inside the parentheses: \[ -2(x^2 - 4x + 4 - 4 - 2) \] \[ -2((x - 2)^2 - 8) \] So the expression under the square root becomes: \[ \sqrt{-2((x - 2)^2 - 8)} \]Complete the Square: Now we can rewrite the integral with the completed square: \[ \int \frac{dx}{\sqrt{-2((x - 2)^2 - 8)}} \] To simplify the integral further, we can factor out the -2 under the square root: \[ \int \frac{dx}{\sqrt{2} \sqrt{8 - (x - 2)^2}} \]Rewrite Integral: Next, we make a trigonometric substitution. We let $ x - 2 = 2\sqrt{2} \sin(\theta) $, so that $ dx = 2\sqrt{2} \cos(\theta) d\theta $ and $ 8 - (x - 2)^2 = 8 - (2\sqrt{2} \sin(\theta))^2 $. Substituting these into the integral, we get: \[ \int \frac{2\sqrt{2} \cos(\theta) d\theta}{\sqrt{2} \sqrt{8 - (2\sqrt{2} \sin(\theta))^2}} \] \[ \int \frac{2\sqrt{2} \cos(\theta) d\theta}{\sqrt{2} \sqrt{8 - 8\sin^2(\theta)}} \] \[ \int \frac{2\sqrt{2} \cos(\theta) d\theta}{\sqrt{2} \sqrt{8(1 - \sin^2(\theta))}} \] \[ \int \frac{2\sqrt{2} \cos(\theta) d\theta}{\sqrt{2} \sqrt{8\cos^2(\theta)}} \] \[ \int \frac{2\sqrt{2} \cos(\theta) d\theta}{2\cos(\theta)} \] \[ \int d\theta \]Trigonometric Substitution: The integral of $ d\theta $ is simply $ \theta $. So we have: \[ \theta + C \] Now we need to express $ \theta $ back in terms of $ x $. From our substitution, we have $ \sin(\theta) = \frac{x - 2}{2\sqrt{2}} $. Therefore, $ \theta = \arcsin\left(\frac{x - 2}{2\sqrt{2}}\right) $. So the indefinite integral is: \[ \arcsin\left(\frac{x - 2}{2\sqrt{2}}\right) + C \]

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Evaluate definite integrals using the chain rule

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\int \frac{dx}{\sqrt{-2x^{2}+8x+4}}

Given Integral: We are given the integral to evaluate: $\newline$ $\int \frac{dx}{\sqrt{-2x^2 + 8x + 4}}$ $\newline$ First, we need to complete the square for the quadratic expression under the square root to simplify the integral. $\newline$ The quadratic expression is $-2$ x^ $2$ + $8$ x + $4$ . To complete the square, we factor out the coefficient of x^ $2$ , which is $-2$ : $\newline$ $-2(x^2 - 4x - 2)$ $\newline$ Now, we find the number to complete the square, which is $\left(\frac{4}{2}\right)^2 = 4$ . We add and subtract this number inside the parentheses: $\newline$ $-2(x^2 - 4x + 4 - 4 - 2)$ $\newline$ $-2((x - 2)^2 - 8)$ $\newline$ So the expression under the square root becomes: $\newline$ $\sqrt{-2((x - 2)^2 - 8)}$
Complete the Square: Now we can rewrite the integral with the completed square: $\newline$ $\int \frac{dx}{\sqrt{-2((x - 2)^2 - 8)}}$ $\newline$ To simplify the integral further, we can factor out the $-2$ under the square root: $\newline$ $\int \frac{dx}{\sqrt{2} \sqrt{8 - (x - 2)^2}}$
Rewrite Integral: Next, we make a trigonometric substitution. We let $x - 2 = 2\sqrt{2} \sin(\theta)$ , so that $dx = 2\sqrt{2} \cos(\theta) d\theta$ and $8 - (x - 2)^2 = 8 - (2\sqrt{2} \sin(\theta))^2$ . $\newline$ Substituting these into the integral, we get: $\newline$ $\int \frac{2\sqrt{2} \cos(\theta) d\theta}{\sqrt{2} \sqrt{8 - (2\sqrt{2} \sin(\theta))^2}}$ $\newline$ $\int \frac{2\sqrt{2} \cos(\theta) d\theta}{\sqrt{2} \sqrt{8 - 8\sin^2(\theta)}}$ $\newline$ $\int \frac{2\sqrt{2} \cos(\theta) d\theta}{\sqrt{2} \sqrt{8(1 - \sin^2(\theta))}}$ $\newline$ $\int \frac{2\sqrt{2} \cos(\theta) d\theta}{\sqrt{2} \sqrt{8\cos^2(\theta)}}$ $\newline$ $\int \frac{2\sqrt{2} \cos(\theta) d\theta}{2\cos(\theta)}$ $\newline$ $\int d\theta$
Trigonometric Substitution: The integral of $d\theta$ is simply $\theta$ . So we have: $\newline$ $\theta + C$ $\newline$ Now we need to express $\theta$ back in terms of $x$ . From our substitution, we have $\sin(\theta) = \frac{x - 2}{2\sqrt{2}}$ . Therefore, $\theta = \arcsin\left(\frac{x - 2}{2\sqrt{2}}\right)$ . $\newline$ So the indefinite integral is: $\newline$ $\arcsin\left(\frac{x - 2}{2\sqrt{2}}\right) + C$