求定积分根号(5-4x-x^2)dx

Question

求定积分 根号(5-4x-x^2)dx

Answer 1

∫√(5 - 4x - x^2) dx
=∫√[9 - (4 + 4x + x^2)] dx
=∫√[9-(2+x)^2] dx
设 x + 2 = 3*sint,则 dx = 3*cost * dt
则原式变为：
= ∫√[9 - 9(sint)^2] *3cost*dt
=∫√(3*cost)^2 * 3cost *dt
=9*∫(cost)^2 dt
=9/2 *∫[2(cost)^2]*dt
=9/2 *∫[1 + cos(2t)] *dt
=9/2 * [ t + 1/2*sin(2t)] + C
=9/2 * t + 9/4 *sin(2t) + C
=9/2 * arcsin[(x+2)/3] + 9/4 * 2*sint * cost + C
=9/2 * arcsin[(x+2)/3] + 9/2 * (x+2)/3 * √{1-[(x+2)/3]^2} + C
=9/2*arcsin[(x+2)/3] + (x+2)/2 * √[9 - (x+2)^2] + C
=9/2*arcsin[(x+2)/3] + (x+2)*√(5-4x-x^2) /2 + C