Question

数学分析，不定积分

数学分析，不定积分这两个计算具体步骤

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Answer 1

第一题

Answer 2

令x=asinu,√(a²-x²)=acosu,则dx=acosudu
原式=∫ a²cos²udu
=a²/2∫ (1+cos2u)du
=a²/2(u+1/2sin2u)+C
=a²u/2+(a²/2)sinucosu+C
由x=asinu,√(a²-x²)=acosu,得：sinu=x/a,u=arcsin(x/a),cosu=√(a²-x²)/a
=(a²/2)arcsin(x/a)+(1/2)x√(a²-x²)+C

追答

令x = a * tanz,dx = a * sec²z dz
√(a² + x²) = a * secz
∫ √(a² + x²) dx = ∫ (a * secz)(a * sec²z) dz
= a²∫ sec³z dz = a² * K
K = ∫ sec³z dz = ∫ secz * sec²z = ∫ secz dtanz
= secz * tanz - ∫ tanz dsecz
= secz * tanz - ∫ tanz * (secz * tanz) dz
= secz * tanz - ∫ secz * tan²z dz
= secz * tanz - ∫ secz * (sec²z - 1) dz
= secz * tanz - K + ∫ secz dz
2K = secz * tanz + ∫ secz dz
K = 1/2 * secz * tanz + 1/2 * ln|secz + tanz|
原式= a²/2 * secz * tanz + a²/2 * ln|secz + tanz|
= (a²/2) * (x/a) * √(a² + x²)/a + (a²/2) * ln|(x/a) + √(a²+x²)/a| + C
= (x/2)√(a² + x²) + (a²/2)ln|x + √(x² + a²)| + C

令x = a * secz,dx = a * secztanz dz
∫ √(x² - a²) dx
= ∫ √(a²sec²z - a²) * (a * secztanz dz)
= a²∫ tan²z * secz dz
= a²∫ (sec²z - 1) * secz dz
= a²∫ sec³z dz - a²∫ secz dz
= a²M - a²N
M = ∫ sec³z dz = ∫ secz dtanz
= secztanz - ∫ tanz dsecz
= secztanz - ∫ tanz * (secztanz dz)
= secztanz - ∫ (sec²z - 1) * secz dz
= secztanz - M + N
2M = secztanz + N => N = (1/2)secztanz + N/2
原式= (a²/2)secztanz + a²N/2 - a²N
= (a²/2)secztanz - (a²/2)∫ secz dz
= (a²/2)secztanz - (a²/2)ln|secz + tanz| + C
= (a²/2)(x/a)[√(x² -a²)/a] - (a²/2)ln|x/a + √(x² - a²)/a| + C
= (x/2)√(x² - a²) - (a²/2)ln|x + √(x² - a²)| + C