Question

请教一道定积分题

Answer 1

∫1/[x+√(a²-x²)]*dx
设x=asint，则dx=acost*dt
sint=x/a，cost=√(1-x²/a²)，t=arcsin(x/a)

原式=∫1/[asint+√(a²-a²sin²t)]*acost*dt
=∫1/(asint+acost)*acost*dt
=∫cost/(sint+cost)*dt
=∫(1/2)[(sint+cost)+(cost-sint)]/(sint+cost)*dt
=∫(1/2)[1+(cost-sint)/(sint+cost)]*dt
=(1/2)∫dt + (1/2)∫(cost-sint)/(sint+cost)*dt
=(1/2)t + (1/2)∫1/(sint+cost)*d(sint+cost)
=(1/2)t + (1/2)ln|sint+cost| + C
=(1/2)arcsin(x/a) + (1/2)ln|x/a+√(1-x²/a²)| + C

∫(0→a)1/[x+√(a²-x²)]*dx
=[ (1/2)arcsin(x/a) + (1/2)ln|x/a+√(1-x²/a²)| ]|(0→a)
=[ (1/2)arcsin(x/a) + (1/2)ln[x/a+√(1-x²/a²)] ]|(0→a)
=[(1/2)*π/2 + (1/2)ln(1+0)] - [(1/2)*0 + (1/2)ln(0+1)]
=π/4