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Help me solve the damn integral

VenZ

Делаем без секанса, только через cos и производную tg.

I = ∫_0^{π/6} tg^2(2x) dx
= ∫_0^{π/6} (1/cos^2(2x) - 1) dx // т.к. tg^2 = (1 - cos^2)/cos^2 = 1/cos^2 - 1

Заметим: d/dx [tg(2x)] = 2/cos^2(2x) ⇒ 1/cos^2(2x) = (1/2) d/dx[tg(2x)].

Тогда
I = ∫_0^{π/6} [(1/2) d/dx tg(2x) - 1] dx
= [ (1/2) tg(2x) - x ]_0^{π/6}
= (1/2) tg(π/3) - π/6
= (1/2)·√3 - π/6.

Code

I = ∫_0^{π/6} tg^2(2x) dx

   = ∫_0^{π/6} (1/cos^2(2x) - 1) dx      // т.к. tg^2 = (1 - cos^2)/cos^2 = 1/cos^2 - 1



Заметим: d/dx [tg(2x)] = 2/cos^2(2x) ⇒ 1/cos^2(2x) = (1/2) d/dx[tg(2x)].



Тогда

I = ∫_0^{π/6} [(1/2) d/dx tg(2x) - 1] dx

  = [ (1/2) tg(2x) - x ]_0^{π/6}

  = (1/2) tg(π/3) - π/6

  = (1/2)·√3 - π/6.

Ответ: √3/2 − π/6 ≈ 0.34243.