Let I = ₀∫^π/2 1/(1 + tan⁷ x) dx 

=> I = ₀∫^π/2 1/(1 + sin⁷ x/cos⁷ x) dx

=> I = ₀∫^π/2 1/{(cos⁷ x + sin⁷ x)/cos⁷ x} dx

=> I = ₀∫^π/2 cos⁷ x/(cos⁷ x + sin⁷ x) dx        ....................1  

=> I = ₀∫^π/2 cos(π/2 - x)⁷ /{cos(π/2 - x)⁷ + sin(π/2 - x)⁷ }dx      {Apply integration rule}

=> I = ₀∫^π/2 sin⁷ x/(cos⁷ x + sin⁷ x)dx   ..............2               

Adding 1 and 2, we get

=> 2I = ₀∫^π/2 (cos⁷ x + sin⁷ x)/(cos⁷ x + sin⁷ x)dx

=> 2I = ₀∫^π/2 dx

=> 2I = [x ₀]^π/2

=> 2I = π/2

=> I = π/(2*2)

=> I = π/4

So, ₀∫^π/2 1/(1 + tan⁷ x) dx = π/4