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Question: $\displaystyle I = \int_{0}^{\pi} \tan^{-1}[\sec2x + \tan2x] \ dx$...

I=∫0πtan⁡−1[sec⁡2x+tan⁡2x] dx\displaystyle I = \int_{0}^{\pi} \tan^{-1}[\sec2x + \tan2x] \ dx

Answer

0

Explanation

Solution

To evaluate the integral I=∫0πtan⁡−1[sec⁡2x+tan⁡2x] dxI = \int_{0}^{\pi} \tan^{-1}[\sec2x + \tan2x] \ dx, we first simplify the expression inside the tan⁡−1\tan^{-1} function.

  1. Simplify the argument of tan⁡−1\tan^{-1}: We have sec⁡2x+tan⁡2x=1cos⁡2x+sin⁡2xcos⁡2x=1+sin⁡2xcos⁡2x\sec2x + \tan2x = \frac{1}{\cos2x} + \frac{\sin2x}{\cos2x} = \frac{1+\sin2x}{\cos2x}. Using the identities 1+sin⁡2x=cos⁡2x+sin⁡2x+2sin⁡xcos⁡x=(cos⁡x+sin⁡x)21+\sin2x = \cos^2x + \sin^2x + 2\sin x \cos x = (\cos x + \sin x)^2 and cos⁡2x=cos⁡2x−sin⁡2x=(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)\cos2x = \cos^2x - \sin^2x = (\cos x - \sin x)(\cos x + \sin x): 1+sin⁡2xcos⁡2x=(cos⁡x+sin⁡x)2(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)=cos⁡x+sin⁡xcos⁡x−sin⁡x\frac{1+\sin2x}{\cos2x} = \frac{(\cos x + \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)} = \frac{\cos x + \sin x}{\cos x - \sin x}. Now, divide the numerator and denominator by cos⁡x\cos x: cos⁡xcos⁡x+sin⁡xcos⁡xcos⁡xcos⁡x−sin⁡xcos⁡x=1+tan⁡x1−tan⁡x\frac{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}} = \frac{1+\tan x}{1-\tan x}. This is the tangent addition formula: tan⁡(π4+x)=tan⁡(π4)+tan⁡x1−tan⁡(π4)tan⁡x=1+tan⁡x1−tan⁡x\tan(\frac{\pi}{4}+x) = \frac{\tan(\frac{\pi}{4})+\tan x}{1-\tan(\frac{\pi}{4})\tan x} = \frac{1+\tan x}{1-\tan x}. So, sec⁡2x+tan⁡2x=tan⁡(π4+x)\sec2x + \tan2x = \tan(\frac{\pi}{4}+x).

  2. Analyze the integrand tan⁡−1(tan⁡(π4+x))\tan^{-1}(\tan(\frac{\pi}{4}+x)): The identity tan⁡−1(tan⁡θ)=θ\tan^{-1}(\tan\theta) = \theta is only valid for θ∈(−π2,π2)\theta \in (-\frac{\pi}{2}, \frac{\pi}{2}). For x∈[0,π]x \in [0, \pi], the argument π4+x\frac{\pi}{4}+x ranges from π4\frac{\pi}{4} to 5π4\frac{5\pi}{4}. We need to define the integrand piecewise:

    • If π4+x∈[π4,π2)\frac{\pi}{4}+x \in [\frac{\pi}{4}, \frac{\pi}{2}), i.e., x∈[0,π4)x \in [0, \frac{\pi}{4}), then tan⁡−1(tan⁡(π4+x))=π4+x\tan^{-1}(\tan(\frac{\pi}{4}+x)) = \frac{\pi}{4}+x.

    • If π4+x∈(π2,5π4]\frac{\pi}{4}+x \in (\frac{\pi}{2}, \frac{5\pi}{4}], i.e., x∈(π4,π]x \in (\frac{\pi}{4}, \pi], then tan⁡−1(tan⁡(π4+x))=(π4+x)−π=x−3π4\tan^{-1}(\tan(\frac{\pi}{4}+x)) = (\frac{\pi}{4}+x) - \pi = x - \frac{3\pi}{4}. This covers x∈(π4,π]x \in (\frac{\pi}{4}, \pi] since π<5π4\pi < \frac{5\pi}{4}.

    The expression sec⁡2x+tan⁡2x\sec2x + \tan2x is undefined when cos⁡2x=0\cos2x=0, which occurs at 2x=π2,3π2,…2x = \frac{\pi}{2}, \frac{3\pi}{2}, \dots. For x∈[0,π]x \in [0, \pi], these points are x=π4x = \frac{\pi}{4} and x=3π4x = \frac{3\pi}{4}. These are points of discontinuity for the integrand. Therefore, the integral is an improper integral. We split the integral at these points.

    The integrand f(x)=tan⁡−1[sec⁡2x+tan⁡2x]f(x) = \tan^{-1}[\sec2x + \tan2x] is: f(x)={π4+xfor x∈[0,π4)x−3π4for x∈(π4,π]f(x) = \begin{cases} \frac{\pi}{4}+x & \text{for } x \in [0, \frac{\pi}{4}) \\ x - \frac{3\pi}{4} & \text{for } x \in (\frac{\pi}{4}, \pi] \end{cases}

  3. Evaluate the improper integral: I=∫0πf(x)dx=lim⁡ϵ1→0+∫0π4−ϵ1(π4+x)dx+lim⁡ϵ2→0+,ϵ3→0+∫π4+ϵ23π4−ϵ3(x−3π4)dx+lim⁡ϵ4→0+∫3π4+ϵ4π(x−3π4)dxI = \int_{0}^{\pi} f(x) dx = \lim_{\epsilon_1 \to 0^+} \int_{0}^{\frac{\pi}{4}-\epsilon_1} (\frac{\pi}{4}+x) dx + \lim_{\epsilon_2 \to 0^+, \epsilon_3 \to 0^+} \int_{\frac{\pi}{4}+\epsilon_2}^{\frac{3\pi}{4}-\epsilon_3} (x-\frac{3\pi}{4}) dx + \lim_{\epsilon_4 \to 0^+} \int_{\frac{3\pi}{4}+\epsilon_4}^{\pi} (x-\frac{3\pi}{4}) dx.

    Let's calculate each part:

    • ∫0π4(π4+x)dx=[π4x+x22]0π4=(π4⋅π4+12(π4)2)−0=π216+π232=2π2+π232=3π232\int_{0}^{\frac{\pi}{4}} (\frac{\pi}{4}+x) dx = \left[\frac{\pi}{4}x + \frac{x^2}{2}\right]_{0}^{\frac{\pi}{4}} = \left(\frac{\pi}{4}\cdot\frac{\pi}{4} + \frac{1}{2}\left(\frac{\pi}{4}\right)^2\right) - 0 = \frac{\pi^2}{16} + \frac{\pi^2}{32} = \frac{2\pi^2+\pi^2}{32} = \frac{3\pi^2}{32}.

    • ∫π43π4(x−3π4)dx=[x22−3π4x]π43π4\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} (x-\frac{3\pi}{4}) dx = \left[\frac{x^2}{2} - \frac{3\pi}{4}x\right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} =(12(3π4)2−3π4(3π4))−(12(π4)2−3π4(π4))= \left(\frac{1}{2}\left(\frac{3\pi}{4}\right)^2 - \frac{3\pi}{4}\left(\frac{3\pi}{4}\right)\right) - \left(\frac{1}{2}\left(\frac{\pi}{4}\right)^2 - \frac{3\pi}{4}\left(\frac{\pi}{4}\right)\right) =(9π232−9π216)−(π232−3π216)= \left(\frac{9\pi^2}{32} - \frac{9\pi^2}{16}\right) - \left(\frac{\pi^2}{32} - \frac{3\pi^2}{16}\right) =(9π2−18π232)−(π2−6π232)= \left(\frac{9\pi^2 - 18\pi^2}{32}\right) - \left(\frac{\pi^2 - 6\pi^2}{32}\right) =−9π232−−5π232=−9π2+5π232=−4π232=−π28= \frac{-9\pi^2}{32} - \frac{-5\pi^2}{32} = \frac{-9\pi^2+5\pi^2}{32} = \frac{-4\pi^2}{32} = -\frac{\pi^2}{8}.

    • ∫3π4π(x−3π4)dx=[x22−3π4x]3π4π\int_{\frac{3\pi}{4}}^{\pi} (x-\frac{3\pi}{4}) dx = \left[\frac{x^2}{2} - \frac{3\pi}{4}x\right]_{\frac{3\pi}{4}}^{\pi} =(π22−3π4π)−(12(3π4)2−3π4(3π4))= \left(\frac{\pi^2}{2} - \frac{3\pi}{4}\pi\right) - \left(\frac{1}{2}\left(\frac{3\pi}{4}\right)^2 - \frac{3\pi}{4}\left(\frac{3\pi}{4}\right)\right) =(2π2−3π24)−(9π232−9π216)= \left(\frac{2\pi^2 - 3\pi^2}{4}\right) - \left(\frac{9\pi^2}{32} - \frac{9\pi^2}{16}\right) =−π24−(9π2−18π232)= -\frac{\pi^2}{4} - \left(\frac{9\pi^2 - 18\pi^2}{32}\right) =−π24−−9π232=−π24+9π232=−8π2+9π232=π232= -\frac{\pi^2}{4} - \frac{-9\pi^2}{32} = -\frac{\pi^2}{4} + \frac{9\pi^2}{32} = \frac{-8\pi^2+9\pi^2}{32} = \frac{\pi^2}{32}.

    Summing these parts: I=3π232−π28+π232=3π2−4π2+π232=032=0I = \frac{3\pi^2}{32} - \frac{\pi^2}{8} + \frac{\pi^2}{32} = \frac{3\pi^2 - 4\pi^2 + \pi^2}{32} = \frac{0}{32} = 0.

The final answer is 0\boxed{0}.

Explanation of the solution:

  1. Simplify the argument of tan⁡−1\tan^{-1}: sec⁡2x+tan⁡2x=1+sin⁡2xcos⁡2x=(cos⁡x+sin⁡x)2(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)=cos⁡x+sin⁡xcos⁡x−sin⁡x=1+tan⁡x1−tan⁡x=tan⁡(π4+x)\sec2x + \tan2x = \frac{1+\sin2x}{\cos2x} = \frac{(\cos x + \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)} = \frac{\cos x + \sin x}{\cos x - \sin x} = \frac{1+\tan x}{1-\tan x} = \tan(\frac{\pi}{4}+x).

  2. Determine the piecewise definition of the integrand f(x)=tan⁡−1(tan⁡(π4+x))f(x) = \tan^{-1}(\tan(\frac{\pi}{4}+x)) over [0,π][0, \pi].

    • For x∈[0,π4)x \in [0, \frac{\pi}{4}), π4+x∈[π4,π2)\frac{\pi}{4}+x \in [\frac{\pi}{4}, \frac{\pi}{2}), so f(x)=π4+xf(x) = \frac{\pi}{4}+x.

    • For x∈(π4,π]x \in (\frac{\pi}{4}, \pi], π4+x∈(π2,5π4]\frac{\pi}{4}+x \in (\frac{\pi}{2}, \frac{5\pi}{4}], so f(x)=(π4+x)−π=x−3π4f(x) = (\frac{\pi}{4}+x) - \pi = x - \frac{3\pi}{4}.

  3. The integral is improper due to discontinuities at x=π4x=\frac{\pi}{4} and x=3π4x=\frac{3\pi}{4}. Split the integral into three parts: I=∫0π4(π4+x)dx+∫π43π4(x−3π4)dx+∫3π4π(x−3π4)dxI = \int_{0}^{\frac{\pi}{4}} (\frac{\pi}{4}+x) dx + \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} (x-\frac{3\pi}{4}) dx + \int_{\frac{3\pi}{4}}^{\pi} (x-\frac{3\pi}{4}) dx.

  4. Evaluate each definite integral:

    • ∫0π4(π4+x)dx=3π232\int_{0}^{\frac{\pi}{4}} (\frac{\pi}{4}+x) dx = \frac{3\pi^2}{32}.

    • ∫π43π4(x−3π4)dx=−π28\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} (x-\frac{3\pi}{4}) dx = -\frac{\pi^2}{8}.

    • ∫3π4π(x−3π4)dx=π232\int_{\frac{3\pi}{4}}^{\pi} (x-\frac{3\pi}{4}) dx = \frac{\pi^2}{32}.

  5. Sum the results: I=3π232−π28+π232=3π2−4π2+π232=0I = \frac{3\pi^2}{32} - \frac{\pi^2}{8} + \frac{\pi^2}{32} = \frac{3\pi^2 - 4\pi^2 + \pi^2}{32} = 0.