1/2 a integration 1/4 pi, 0 (1+tan^2 theta)/(1+tan theta + tan^2 theta) *d theta. use substitution t = tan theta to find exact value of area

Question

1/2 a integration 1/4 pi, 0  (1+tan^2 theta)/(1+tan theta + tan^2 theta)  *d theta. use substitution t = tan theta to find exact value of area

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the exact value of the area, we start with the integral: $\frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{1 + \tan^2 \theta}{1 + \tan \theta + \tan^2 \theta} \, d\theta$. We will use the substitution $t = \tan \theta$, which gives us $d\theta = \frac{1}{1+t^2} \, dt$. ‖ ‖ step 2 ⋮ Changing the limits of integration: when $\theta = 0$, $t = \tan(0) = 0$; when $\theta = \frac{\pi}{4}$, $t = \tan\left(\frac{\pi}{4}\right) = 1$. Thus, the integral becomes: $\frac{1}{2} \int_0^1 \frac{1 + t^2}{1 + t + t^2} \cdot \frac{1}{1+t^2} \, dt$. ‖ ‖ step 3 ⋮ Simplifying the integrand: $\frac{1 + t^2}{1 + t + t^2} \cdot \frac{1}{1+t^2} = \frac{1}{1 + t + t^2}$. Therefore, the integral simplifies to: $\frac{1}{2} \int_0^1 \frac{1}{1 + t + t^2} \, dt$. ‖ ‖ step 4 ⋮ To evaluate the integral $\int_0^1 \frac{1}{1 + t + t^2} \, dt$, we can use partial fraction decomposition or recognize it as a standard integral. The result of this integral is $\frac{\pi}{3\sqrt{3}}$. Thus, the area is: $\frac{1}{2} \cdot \frac{\pi}{3\sqrt{3}} = \frac{\pi}{6\sqrt{3}}$. ‖ ∻Answer∻ ⚹ The exact value of the area is $\frac{\pi}{6\sqrt{3}}$ ⚹ ∻Key Concept∻ ⚹ The use of substitution in integrals allows for simplification and easier evaluation of the integral. ⚹ ∻Explanation∻ ⚹ By substituting $t = \tan \theta$, we transformed the integral into a more manageable form, leading to the exact area calculation. ⚹◊What is the derivative of the function $ t = \tan(\theta) $ with respect to $ \theta $?⍭ Generate me a similar question◊

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