int_(-(pi)/(6))^((pi)/(3))(-sin x)/(cos^(2)x)dx Click to view the table of general integ

Identify Integral: Identify the integral to solve: ∫ from -π/6 to π/3 of (-sin x)/(cos^2 x) dx. Recognize this as a standard integral involving trigonometric identities.Use Substitution: Use substitution: Let u = cos(x), then du = -sin(x) dx. Rewrite the integral in terms of u: ∫ of (-sin x)/cos^2(x) dx = ∫ du/u^2.Rewrite Integral: Calculate the new limits of integration: When x = -π/6, cos(-π/6) = cos(π/6) = √3/2. When x = π/3, cos(π/3) = 1/2. So, the limits change from u = √3/2 to u = 1/2.Calculate New Limits: Evaluate the integral: ∫ du/u^2 from √3/2 to 1/2 = [-1/u] from √3/2 to 1/2. Calculate: [-1/(1/2)] - [-1/(√3/2)] = -2 + 2/√3. Simplify: -2 + 2/√3 = (-6 + 2√3)/3.

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int_(-(pi)/(6))^((pi)/(3))(-sin x)/(cos^(2)x)dx
Click to view the table of general integ

$\int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{-\sin x}{\cos ^{2} x} d x$ $\newline$

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Calculus

Evaluate definite integrals using the chain rule

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\int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{-\sin x}{\cos ^{2} x} d x

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Identify Integral: Identify the integral to solve: $\int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{-\sin x}{\cos^2 x} \, dx$ . Recognize this as a standard integral involving trigonometric identities.
Use Substitution: Use substitution: $\newline$ Let $u = \cos(x)$ , then $du = -\sin(x) dx$ . $\newline$ Rewrite the integral in terms of $u$ : $\newline$ $\int \frac{-\sin x}{\cos^2(x)} dx = \int \frac{du}{u^2}$ .
Rewrite Integral: Calculate the new limits of integration: $\newline$ When $x = -\frac{\pi}{6}$ , $\cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \sqrt{3}/2$ . $\newline$ When $x = \frac{\pi}{3}$ , $\cos(\frac{\pi}{3}) = 1/2$ . $\newline$ So, the limits change from $u = \sqrt{3}/2$ to $u = 1/2$ .
Calculate New Limits: Evaluate the integral: $\int \frac{du}{u^2}$ from $\sqrt{3}/2$ to $1/2$ = $[-1/u]$ from $\sqrt{3}/2$ to $1/2$ . Calculate: $[-1/(1/2)] - [-1/(\sqrt{3}/2)]$ = $-2 + 2/\sqrt{3}$ . Simplify: $-2 + 2/\sqrt{3}$ = $(-6 + 2\sqrt{3})/3$ .