int(cos^(3)xdx)/(sin^(2)x-4)

Identify Integral: Identify the integral to be solved. We need to evaluate the integral of the function (cos^3(x))/(sin^2(x)-4) with respect to x.Simplify if Possible: Simplify the integral if possible. The integral does not simplify easily due to the complex relationship between the numerator and the denominator. We may need to consider a substitution that relates sin(x) and cos(x) or use trigonometric identities to simplify the expression.Consider Substitution: Consider a substitution. Let u = sin(x), then du = cos(x)dx. This substitution will help us rewrite the integral in terms of u, which might simplify the expression.Rewrite Using Substitution: Rewrite the integral using the substitution. The integral becomes ∫(cos^2(x)cos(x)dx)/(u^2-4). Since we have du = cos(x)dx, we can replace cos(x)dx with du. However, we still have a cos^2(x) term in the numerator, which we need to express in terms of u.Use Trig Identity: Use a trigonometric identity to express cos^2(x) in terms of u. We know that sin^2(x) + cos^2(x) = 1, so cos^2(x) = 1 - sin^2(x). Substituting u = sin(x), we get cos^2(x) = 1 - u^2.Substitute with u: Substitute cos^2(x) with 1 - u^2 in the integral. The integral now becomes ∫((1 - u^2)du)/(u^2-4).Simplify Integral: Simplify the integral. We can now integrate term by term. The integral becomes ∫(du)/(u^2-4) - ∫(u^2du)/(u^2-4).Evaluate First Term: Evaluate the first term of the integral. The first term ∫(du)/(u^2-4) is a standard integral that can be solved using partial fractions. However, I realize there is a mistake in the previous steps. We have not accounted for the cos^3(x) correctly in our substitution. We only replaced cos(x)dx with du, but we still have an extra cos^2(x) term that needs to be accounted for. This is a math error.

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$\int \frac{\cos ^{3} x d x}{\sin ^{2} x-4}$

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Calculus

Evaluate definite integrals using the chain rule

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\int \frac{\cos ^{3} x d x}{\sin ^{2} x-4}

Identify Integral: Identify the integral to be solved. $\newline$ We need to evaluate the integral of the function $\frac{\cos^3(x)}{\sin^2(x)-4}$ with respect to $x$ .
Simplify if Possible: Simplify the integral if possible. $\newline$ The integral does not simplify easily due to the complex relationship between the numerator and the denominator. We may need to consider a substitution that relates $\sin(x)$ and $\cos(x)$ or use trigonometric identities to simplify the expression.
Consider Substitution: Consider a substitution. $\newline$ Let $u = \sin(x)$ , then $du = \cos(x)dx$ . This substitution will help us rewrite the integral in terms of $u$ , which might simplify the expression.
Rewrite Using Substitution: Rewrite the integral using the substitution. $\newline$ The integral becomes $\int\frac{\cos^2(x)\cos(x)\,dx}{u^2-4}$ . Since we have $du = \cos(x)\,dx$ , we can replace $\cos(x)\,dx$ with $du$ . However, we still have a $\cos^2(x)$ term in the numerator, which we need to express in terms of $u$ .
Use Trig Identity: Use a trigonometric identity to express $\cos^2(x)$ in terms of $u$ . We know that $\sin^2(x) + \cos^2(x) = 1$ , so $\cos^2(x) = 1 - \sin^2(x)$ . Substituting $u = \sin(x)$ , we get $\cos^2(x) = 1 - u^2$ .
Substitute with $u$ : Substitute $\cos^2(x)$ with $1 - u^2$ in the integral. $\newline$ The integral now becomes $\int\frac{(1 - u^2)\,du}{u^2-4}$ .
Simplify Integral: Simplify the integral. $\newline$ We can now integrate term by term. The integral becomes $\int\frac{du}{u^2-4}$ - $\int\frac{u^2du}{u^2-4}$ .
Evaluate First Term: Evaluate the first term of the integral. $\newline$ The first term $\int\frac{du}{u^2-4}$ is a standard integral that can be solved using partial fractions. However, I realize there is a mistake in the previous steps. We have not accounted for the $\cos^3(x)$ correctly in our substitution. We only replaced $\cos(x)dx$ with $du$ , but we still have an extra $\cos^2(x)$ term that needs to be accounted for. This is a math error.