int_((pi)/(6))^((pi)/(3))((1)/(cos^(2)x)-(1)/(sin^(2)x))dx 0,5 0

Simplify integrand: Step 1: Simplify the integrand. We start by recognizing that 1/cos^2(x) is sec^2(x) and 1/sin^2(x) is csc^2(x). So, the integral becomes: ∫(sec^2(x) - csc^2(x)) dxSeparate the integral: Step 2: Separate the integral. The integral of sec^2(x) is tan(x), and the integral of csc^2(x) is -cot(x). Therefore: ∫sec^2(x) dx - ∫csc^2(x) dx = tan(x) + cot(x)Evaluate limits: Step 3: Evaluate from π/6 to π/3. Plug in the limits of integration: [tan(x) + cot(x)] evaluated from π/6 to π/3 = [tan(π/3) + cot(π/3)] - [tan(π/6) + cot(π/6)]Calculate values: Step 4: Calculate the values. tan(π/3) = √3, cot(π/3) = 1/√3, tan(π/6) = 1/√3, cot(π/6) = √3 So, [√3 + 1/√3] - [1/√3 + √3] = 0

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int_((pi)/(6))^((pi)/(3))((1)/(cos^(2)x)-(1)/(sin^(2)x))dx
0,5
0

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\frac{1}{\cos ^{2} x}-\frac{1}{\sin ^{2} x}\right) d x$ $\newline$ $0$ , $5$ $\newline$ $0$

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Calculus

Evaluate definite integrals using the chain rule

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\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\frac{1}{\cos ^{2} x}-\frac{1}{\sin ^{2} x}\right) d x

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0

5

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0

Simplify integrand: Step $1$ : Simplify the integrand. $\newline$ We start by recognizing that $\frac{1}{\cos^2(x)}$ is $\sec^2(x)$ and $\frac{1}{\sin^2(x)}$ is $\csc^2(x)$ . So, the integral becomes: $\newline$ $\int(\sec^2(x) - \csc^2(x)) \, dx$
Separate the integral: Step $2$ : Separate the integral. $\newline$ The integral of $\sec^2(x)$ is $\tan(x)$ , and the integral of $\csc^2(x)$ is $-\cot(x)$ . Therefore: $\newline$ $\int \sec^2(x) \, dx - \int \csc^2(x) \, dx = \tan(x) + \cot(x)$
Evaluate limits: Step $3$ : Evaluate from $\frac{\pi}{6}$ to $\frac{\pi}{3}$ . $\newline$ Plug in the limits of integration: $\newline$ $[\tan(x) + \cot(x)]$ evaluated from $\frac{\pi}{6}$ to $\frac{\pi}{3}$ $\newline$ = $[\tan(\frac{\pi}{3}) + \cot(\frac{\pi}{3})] - [\tan(\frac{\pi}{6}) + \cot(\frac{\pi}{6})]$
Calculate values: Step $4$ : Calculate the values. $\newline$ $\tan(\frac{\pi}{3}) = \sqrt{3}$ , $\cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}$ , $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$ , $\cot(\frac{\pi}{6}) = \sqrt{3}$ $\newline$ So, $[\sqrt{3} + \frac{1}{\sqrt{3}}] - [\frac{1}{\sqrt{3}} + \sqrt{3}] = 0$