0到a根号a^2+x^2的定积分
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利用三角代换来求
∫(0→a) √(a² + x²) dx
令x = atanθ,dx = asec²θ dθ
当x = 0,θ = 0
当x = a,θ = π/4
∫(0→π/4) (asecθ)(asec²θ) dθ
= a²∫(0→π/4) sec³θ dθ
= (a²)(1/2)[secθtanθ + ln(secθ + tanθ)]：[0→π/4]
= (a²/2)[sec(π/4)tan(π/4) + ln(sec(π/4) + tan(π/4)) - (a²/2)[0 + ln(sec(0))]
= a²/√2 + (a²/2)ln(1 + √2)
∫(0→a) √(a² + x²) dx
令x = atanθ,dx = asec²θ dθ
当x = 0,θ = 0
当x = a,θ = π/4
∫(0→π/4) (asecθ)(asec²θ) dθ
= a²∫(0→π/4) sec³θ dθ
= (a²)(1/2)[secθtanθ + ln(secθ + tanθ)]：[0→π/4]
= (a²/2)[sec(π/4)tan(π/4) + ln(sec(π/4) + tan(π/4)) - (a²/2)[0 + ln(sec(0))]
= a²/√2 + (a²/2)ln(1 + √2)
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