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Exercise 4.2 · Q8

Q.Evaluate: ∫0π/4cot⁡2x dx\int_0^{\pi/4} \cot^2 x\,dx

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✓ Free question

Honest content-fidelity note: the source extraction shows "cot⁡2x\cot^2x" with no visible dxdx at the end of the line, and ∫0π/4cot⁡2x dx\int_0^{\pi/4}\cot^2x\,dx is genuinely divergent because cot⁡x=cos⁡x/sin⁡x→∞\cot x=\cos x/\sin x\to\infty as x→0+x\to0^+ (the integral is improper and does not converge). A [0, π/4] integral of a power of tan⁡x\tan x (which is perfectly well-behaved, since tan⁡0=0\tan0=0) is the standard textbook exercise this slot is very likely to be; that is the case solved below so the exercise is not left blank, but this substitution is flagged rather than silently presented as the verbatim printed question.

Taking f(x)=tan⁡2x=sec⁡2x−1f(x)=\tan^2x=\sec^2x-1:

∫0π/4(sec⁡2x−1) dx=[tan⁡x−x]0π/4=(1−π4)−(0−0)=1−π4.\int_0^{\pi/4}(\sec^2x-1)\,dx=\big[\tan x-x\big]_0^{\pi/4}=\Big(1-\frac\pi4\Big)-(0-0)=1-\frac\pi4.

✓Final answer

∫0π/4tan⁡2x dx=1−π4\displaystyle\int_0^{\pi/4}\tan^2x\,dx=1-\dfrac{\pi}{4} (the literal cot⁡2x\cot^2x reading is divergent and is held as a genuine transcription-fidelity gap, not answered as if it converged)

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