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

∫−π6π3−sin⁡xcos⁡2xdx \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{-\sin x}{\cos ^{2} x} d x \newline

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Evaluate definite integrals using the chain ruleright chevron icon

Full solution

Q. ∫−π6π3−sin⁡xcos⁡2xdx \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{-\sin x}{\cos ^{2} x} d x \newline
  1. Identify Integral: Identify the integral to solve: ∫−π6π3−sin⁡xcos⁡2x dx\int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{-\sin x}{\cos^2 x} \, dx. Recognize this as a standard integral involving trigonometric identities.
  2. Use Substitution: Use substitution:\newlineLet u=cos⁡(x)u = \cos(x), then du=−sin⁡(x)dxdu = -\sin(x) dx.\newlineRewrite the integral in terms of uu:\newline∫−sin⁡xcos⁡2(x)dx=∫duu2\int \frac{-\sin x}{\cos^2(x)} dx = \int \frac{du}{u^2}.
  3. Rewrite Integral: Calculate the new limits of integration:\newlineWhen x=−π6x = -\frac{\pi}{6}, cos⁡(−π6)=cos⁡(π6)=3/2\cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \sqrt{3}/2.\newlineWhen x=π3x = \frac{\pi}{3}, cos⁡(π3)=1/2\cos(\frac{\pi}{3}) = 1/2.\newlineSo, the limits change from u=3/2u = \sqrt{3}/2 to u=1/2u = 1/2.
  4. Calculate New Limits: Evaluate the integral: ∫duu2\int \frac{du}{u^2} from 3/2\sqrt{3}/2 to 1/21/2 = [−1/u][-1/u] from 3/2\sqrt{3}/2 to 1/21/2. Calculate: [−1/(1/2)]−[−1/(3/2)][-1/(1/2)] - [-1/(\sqrt{3}/2)] = −2+2/3-2 + 2/\sqrt{3}. Simplify: −2+2/3-2 + 2/\sqrt{3} = (−6+23)/3(-6 + 2\sqrt{3})/3.

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