Question

求∫1/(1+sin^2x)dx的积分，上下线为0到π/2 要详解？？？？？？？

1+sin^2x注意这里是1+sinx的平方 不要看错了，要步骤啊~~~
那位大侠能帮帮小弟~~~~

Answer 1

令tanx = t,x = arctant
则dx = dt/(1+t²)
1+sin²x = 1 + t²/(1+t²)
∫dx/(1+sin²x)
=∫dt/(1+2t²)
=1/√2 ∫d√2t/[1+(√2t)²]
=1/√2 arctan√2t + C
=1/√2 arctan(√2tanx) + C
x = π/2时,1/√2 arctan(√2tanx) = π/2√2
x = 0时,1/√2 arctan(√2tanx) = 0
原式 = π/2√2

Answer 2

1/[1+(sinx)^2]=1/{(1/2)[2+2(sinx)^2]}
=1/{(1/2)[3-cos2x]}
=2/(3-cos2x)
∫[2/(3-cos2x)]
不会啊！
电脑1.1107……