Elementary Calculus: Theorem 2:

THEOREM 2

(i)

(where - 1 < x < 1).

(ii)

(iii)

(where |x| > 1).

PROOF

We prove the first part of (i) and (iii). Since the derivatives exist we may use implicit differentiation.

(i) Let y = arcsin x. Then

x = sin y, -π/2 ≤ y ≤ π/2, dx = cos y dy. From sin² y + cos² y = 1 we get

Since - π/2 ≤ y ≤ π/2, cos y ≥ 0. Then

Substituting,

(iii) Let y = arcsec x.

Then

x = sec y, 0 ≤ y ≤ π,

dx = sec y tan y dy.

From tan² y + 1 = sec² y we get

tan y = =

Since 0 ≤ y ≤ π, tan y and sec have the same sign.

Therefore

sec y tan y ≥ 0

and

When we turn these formulas for derivatives around we get some surprising new integration formulas.

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Theorem 2: THEOREM 2 (i) (where - 1 < x < 1). (where - 1 < x < 1). (ii) (iii) (where \|x\| > 1). (where \|x\| > 1). PROOF We prove the first part of (i) and (iii). Since the derivatives exist we may use implicit differentiation. (i) Let y = arcsin x. Then x = sin y, -π/2 ≤ y ≤ π/2, dx = cos y dy. From sin² y + cos² y = 1 we get Since - π/2 ≤ y ≤ π/2, cos y ≥ 0. Then Substituting, (iii) Let y = arcsec x. Then x = sec y, 0 ≤ y ≤ π, dx = sec y tan y dy. From tan² y + 1 = sec² y we get tan y = = Since 0 ≤ y ≤ π, tan y and sec have the same sign. Therefore sec y tan y ≥ 0 and When we turn these formulas for derivatives around we get some surprising new integration formulas.
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