Difficult integration problem (from my perspective).?

Favorite Answer

a) ∫(x = -π/2 to π/2) tan²x dx/(π²/4 - x²)
= 2 ∫(x = 0 to π/2) tan²x dx/(π²/4 - x²), since the integrand is even
= 2 ∫(x = 0 to π/2) tan²x dx/[(π/2 - x) (π/2 + x)]
= 2 ∫(w=π/2 to 0) tan²(π/2 - w) * (-dw)/[w (π/2 + π/2 - w)], letting w=π/2 - x
= 2 ∫(w = 0 to π/2) tan²(π/2 - w) dw/[w(π - w)]
= 2 ∫(w = 0 to π/2) cot²w dw/[w(π - w)]

Since the integral is only improper at w = 0, this will converge
iff ∫(w = 0 to π/4) cot²w dw/[w(π - w)] converges.

However, for w in (0, π/4), we have
cot²w/[w(π - w)]
> cot²w/[w(π - 0)]
= 1/(πw tan²w)
> 1/(πw tan²π/4)
= 1/(πw).

Since ∫(w = 0 to π/4) dw/(πw) = (1/π) ln w {for w = 0 to π/4} = ∞, we conclude that ∫(w = 0 to π/4) cot²w dw/[w(π - w)] diverges by the Comparison Test.

Therefore, ∫(x = -π/2 to π/2) tan²x dx/(π²/4 - x²) is divergent.
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b) According to Wolfram Alpha,
∫(x = -π/2 to π/2) |tan x| dx/(π²/4 - x²) is divergent.

(This can be showed with similar reasoning as in part a.)
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I hope this helps!...Show more