int(1)/(sqrt(9-4x^(2)))dx

Recognize standard form: Recognize the integral as a standard form. The integral int(1)/(sqrt(9-4x^(2)))dx resembles the integral of the form int(1)/(sqrt(a^2 - x^2))dx, which is a standard form for the inverse trigonometric function arcsin(x/a).Identify a^2: Identify a^2 from the integral. In our integral, a^2 = 9, so a = 3. This means we can rewrite the integral in terms of a standard arcsin function.Use substitution x=(3/2)sin(u): Rewrite the integral using the substitution x = (3/2)sin(u). Let x = (3/2)sin(u), then dx = (3/2)cos(u)du. We need to find the limits of integration in terms of u, but since this is an indefinite integral, we will just perform the substitution.Substitute x and dx: Substitute x and dx in the integral. Substituting x and dx into the integral, we get int(1)/(sqrt(9-4(3/2sin(u))^2)) * (3/2)cos(u)du.Simplify expression: Simplify the integral. Simplify the expression under the square root: 9 - 4(3/2)^2sin^2(u) = 9 - 9sin^2(u) = 9(1 - sin^2(u)) = 9cos^2(u). The integral becomes int(1)/(sqrt(9cos^2(u))) * (3/2)cos(u)du.Further simplify integral: Further simplify the integral. Since sqrt(9cos^2(u)) = 3cos(u), the integral simplifies to int(1)/(3cos(u)) * (3/2)cos(u)du = int(1/2)du.Integrate with respect to u: Integrate with respect to u. The integral of 1/2 with respect to u is (1/2)u + C.Substitute back for u: Substitute back for u. We need to express our answer in terms of x, so we substitute back using the original substitution x = (3/2)sin(u). We solve for u: sin(u) = (2/3)x, so u = arcsin((2/3)x).Write final answer: Write the final answer. The final answer is (1/2)arcsin((2/3)x) + C.

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$\int \frac{1}{\sqrt{9-4 x^{2}}} d x$

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Evaluate definite integrals using the chain rule

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\int \frac{1}{\sqrt{9-4 x^{2}}} d x

Recognize standard form: Recognize the integral as a standard form. The integral $\int\frac{1}{\sqrt{9-4x^{2}}}\,dx$ resembles the integral of the form $\int\frac{1}{\sqrt{a^{2} - x^{2}}}\,dx$ , which is a standard form for the inverse trigonometric function $\arcsin(\frac{x}{a})$ .
Identify $a^2$ : Identify $a^2$ from the integral. $\newline$ In our integral, $a^2 = 9$ , so $a = 3$ . This means we can rewrite the integral in terms of a standard $ext{arcsin}$ function.
Use substitution $x=\frac{3}{2}\sin(u)$ : Rewrite the integral using the substitution $x = \frac{3}{2}\sin(u)$ . Let $x = \frac{3}{2}\sin(u)$ , then $dx = \frac{3}{2}\cos(u)du$ . We need to find the limits of integration in terms of $u$ , but since this is an indefinite integral, we will just perform the substitution.
Substitute $x$ and $dx$ : Substitute $x$ and $dx$ in the integral. $\newline$ Substituting $x$ and $dx$ into the integral, we get $\int \frac{1}{\sqrt{9-4\left(\frac{3}{2}\sin(u)\right)^2}} \cdot \left(\frac{3}{2}\right)\cos(u)\,du$ .
Simplify expression: Simplify the integral. $\newline$ Simplify the expression under the square root: $9 - 4\left(\frac{3}{2}\right)^2\sin^2(u) = 9 - 9\sin^2(u) = 9(1 - \sin^2(u)) = 9\cos^2(u)$ . $\newline$ The integral becomes $\int \frac{1}{\sqrt{9\cos^2(u)}} \cdot \left(\frac{3}{2}\right)\cos(u)\,du$ .
Further simplify integral: Further simplify the integral. $\newline$ Since $\sqrt{9\cos^2(u)} = 3\cos(u)$ , the integral simplifies to $\int\frac{1}{3\cos(u)} \cdot \left(\frac{3}{2}\right)\cos(u)\,du = \int\frac{1}{2}\,du$ .
Integrate with respect to $u$ : Integrate with respect to $u$ . The integral of $\frac{1}{2}$ with respect to $u$ is $\left(\frac{1}{2}\right)u + C$ .
Substitute back for u: Substitute back for u. $\newline$ We need to express our answer in terms of $x$ , so we substitute back using the original substitution $x = \frac{3}{2}\sin(u)$ . We solve for $u$ : $\sin(u) = \frac{2}{3}x$ , so $u = \arcsin(\frac{2}{3}x)$ .
Write final answer: Write the final answer. $\newline$ The final answer is $(\frac{1}{2})\arcsin(\frac{2}{3}x) + C$ .