lanboy520
幼苗
共回答了20个问题采纳率:90% 举报
(1)设x=tant,则dx=sec²tdt
∴原式=∫sec²tdt/sec³t
=∫costdt
=sint+C (C是积分常数)
=x/√(x²+1)+C;
(2)设x=2sint,则dx=2costdt
∴原式=∫(2sint)²(2cost)dt/(2cost)
=4∫sin²tdt
=2∫(1-cos(2t))dt
=2(t-sin(2t)/2)+C (C是积分常数)
=2t-sin(2t)+C
=2arcsin(x/2)-x√(4-x²)/2+C.
1年前
7