求不定积分1题,1/[x+√(1-x∧2)]
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求不定积分1题,
1/[x+√(1-x∧2)]
1/[x+√(1-x∧2)]
![求不定积分1题,1/[x+√(1-x∧2)]](/uploads/image/z/3285790-70-0.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%861%E9%A2%98%2C1%2F%5Bx%2B%E2%88%9A%281-x%E2%88%A72%29%5D)
代换x=sint π/2>t>-π/2 cost>0
dx=costdt
∫1/[x+√(1-x^2)]dx
=∫cost/(sint+cost)dt
I=∫cost/(sint+cost)dt
J=∫sint/(sint+cost)dt
I+J=∫(sint+cost)/(sint+cost)dt=∫dt=t=arcsinx+C
J-I=∫(cost-sint)/(sint+cost)dt=
=∫d(sint+cost)/(sint+cost)dt
=ln(sint+cost)
=ln(x+√(1-x^2))+C
I=1/2*((I+J)-(J-I))
=1/2arcsinx-1/2ln(x+√(1-x^2))+C
dx=costdt
∫1/[x+√(1-x^2)]dx
=∫cost/(sint+cost)dt
I=∫cost/(sint+cost)dt
J=∫sint/(sint+cost)dt
I+J=∫(sint+cost)/(sint+cost)dt=∫dt=t=arcsinx+C
J-I=∫(cost-sint)/(sint+cost)dt=
=∫d(sint+cost)/(sint+cost)dt
=ln(sint+cost)
=ln(x+√(1-x^2))+C
I=1/2*((I+J)-(J-I))
=1/2arcsinx-1/2ln(x+√(1-x^2))+C