$12$ ). $\int \frac{\cos ^{3} x}{\sin ^{2} x-2} d x=$

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Evaluate definite integrals using the power rule

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12

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

Simplify Integral: Let's start by simplifying the integral a bit. We can rewrite the integral as: $\int(\cos^3(x)/\sin^2(x))\,dx - 2 \times \int(\cos^3(x))\,dx$
First Integral Substitution: Now, let's tackle the first integral. We can use a substitution where $u = \sin(x)$ , which means $du = \cos(x)dx$ .
First Integral Simplification: Substituting, we get: $\int\left(\frac{\cos^2(x)}{u^2}\right) \cos(x)\,dx = \int\left(\frac{1-u^2}{u^2}\right)du$
First Integral Integration: This simplifies to: $\int \frac{1}{u^2}du - \int \frac{u^2}{u^2}du = \int \frac{1}{u^2}du - \int 1du$
Second Integral Approach: Now we can integrate: $\int(\frac{1}{u^2})du = -\frac{1}{u}$ and $\int(1)du = u$
Second Integral Substitution: So the first part of the integral becomes: $-\frac{1}{\sin(x)} + \sin(x)$
Second Integral Integration: Now, let's look at the second integral, $2 \times \int(\cos^3(x))\,dx$ . We can use a power-reducing formula for $\cos^2(x) = 1 - \sin^2(x)$ .
Combine Integrals: The integral becomes: $2 \int \cos(x) \cdot (1 - \sin^2(x))\,dx$
Simplify Expression: We can use the substitution $u = \sin(x)$ again, so $du = \cos(x)dx$ .
Correct Simplification: Substituting, we get: $2 \times \int (1 - u^2) \, du$
Correct Simplification: Substituting, we get: $\newline$ $2 \int (1 - u^2) \, du$ Integrating, we find: $\newline$ $2 \left( u - \frac{u^3}{3} \right) + C$
Correct Simplification: Substituting, we get: $\newline$ $2 \int (1 - u^2) \, du$ Integrating, we find: $\newline$ $2 \left( u - \frac{u^3}{3} \right) + C$ Substituting back $\sin(x)$ for $u$ , we get: $\newline$ $2 \left( \sin(x) - \frac{\sin^3(x)}{3} \right) + C$
Correct Simplification: Substituting, we get: $\newline$ $2 \int (1 - u^2) \, du$ Integrating, we find: $\newline$ $2 \cdot (u - \frac{u^3}{3}) + C$ Substituting back $\sin(x)$ for $u$ , we get: $\newline$ $2 \cdot (\sin(x) - \frac{\sin^3(x)}{3}) + C$ Now, let's combine both parts of the integral: $\newline$ $(-\frac{1}{\sin(x)} + \sin(x)) + (2 \cdot (\sin(x) - \frac{\sin^3(x)}{3})) + C$
Correct Simplification: Substituting, we get: $\newline$ $2 \int (1 - u^2) \, du$ Integrating, we find: $\newline$ $2 \left(u - \frac{u^3}{3}\right) + C$ Substituting back $\sin(x)$ for $u$ , we get: $\newline$ $2 \left(\sin(x) - \frac{\sin^3(x)}{3}\right) + C$ Now, let's combine both parts of the integral: $\newline$ $\left(-\frac{1}{\sin(x)} + \sin(x)\right) + \left(2 \left(\sin(x) - \frac{\sin^3(x)}{3}\right)\right) + C$ Simplify the expression: $\newline$ -\frac{$1$}{\sin(x)} + \(3\sin(x) - \left(\frac{ $2$ }{ $3$ }\right)\sin^ $3$ (x) + C
Correct Simplification: Substituting, we get: $\newline$ $2 \int (1 - u^2) \, du$ Integrating, we find: $\newline$ $2 \cdot (u - \frac{u^3}{3}) + C$ Substituting back $\sin(x)$ for $u$ , we get: $\newline$ $2 \cdot (\sin(x) - \frac{\sin^3(x)}{3}) + C$ Now, let's combine both parts of the integral: $\newline$ $(-\frac{1}{\sin(x)} + \sin(x)) + (2 \cdot (\sin(x) - \frac{\sin^3(x)}{3})) + C$ Simplify the expression: $\newline$ $-\frac{1}{\sin(x)} + 3\sin(x) - \frac{2}{3}\sin^3(x) + C$ Oops, we made a mistake in the simplification. The correct simplification should be: $\newline$ $-\frac{1}{\sin(x)} + \sin(x) + 2\sin(x) - \frac{2}{3}\sin^3(x) + C$