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Rewrite sec⁴(x) as sec²(x)sec²(x). The integral is ∫5tan(x)sec²(x)sec²(x) dx.Let u = tan(x). Then du = sec²(x) dx.Also, sec²(x) = 1 + tan²(x) = 1 + u².Substitute these into the integral:∫5u(1 + u²) du = ∫(5u + 5u³) du= 5u²/2 + 5u⁴/4 + C= 5tan²(x)/2 + 5tan⁴(x)/4 + C.Now, evaluate the definite integral from 0 to θ:[5tan²(x)/2 + 5tan⁴(x)/4]0θ = (5tan²(θ)/2 + 5tan⁴(θ)/4) - (5tan²(0)/2 + 5tan⁴(0)/4)= 5tan²(θ)/2 + 5tan⁴(θ)/4 - 0 (since tan(0)=0).We are given that this equals 15/4:5tan²(θ)/2 + 5tan⁴(θ)/4 = 15/4Multiply by 4 to clear denominators: 10tan²(θ) + 5tan⁴(θ) = 15.Divide by 5: 2tan²(θ) + tan⁴(θ) = 3.Let y = tan²(θ). The equation becomes y² + 2y - 3 = 0.Factor the quadratic equation: (y + 3)(y - 1) = 0.So, y = -3 or y = 1. Since y = tan²(θ), it cannot be negative, so tan²(θ) = 1.This means tan(θ) = ±1.Given the condition 0 < θ < π/2, tan(θ) must be positive, so tan(θ) = 1.Therefore, θ = π/4.