If the integrals i 1 = ∫ 0 ∞ 1 1 + x 8 d x and i 2 = ∫ 0 1 1 ( 1 - x 8 ) 1 / 8 d x, then

Question

If the integrals i 1 = ∫ 0 ∞ 1 1 + x 8 d x and i 2 = ∫ 0 1 1 ( 1 - x 8 ) 1 / 8 d x, then, i 1 = 2 i 2, i 2 = 8 i 1, i 1 = i 2, i 1 = 8 i 2

MARKS App · Accepted Answer

In i 1 put x 8 = t a n 2 θ ⇒ 8 x 7 d x = 2 tan ⁡ θ s e c 2 θ d θ ⇒ i 1 = ∫ 0 π / 2 2 tan ⁡ θ s e c 2 θ d θ s e c 2 θ · 8 · t a n 7 / 4 θ ⇒ i 1 = 1 4 ∫ 0 π / 2 t a n - 3 / 4 θ d θ In i 2 put x 8 = s i n 2 θ ⇒ i 2 = ∫ 0 π / 2 2 sin θ cos θ d θ ( 1 - sin 2 θ ) 1 / 8 · 8 sin 7 4 θ = 1 4 ∫ 0 π / 2 t a n - 3 / 4 θ d θ = i 1 ⇒ i 2 = i 1

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