Question

一道定积分的题:)

RT

Answer 1

解：dx/[1+(cosx)^2] = (secx)^2dx/[1+(secx)^2]=d(tanx)/[2+(tanx)^2]=d(tanx)/[(√2)^2+(tanx)^2].
所以，原式=【1/√2 * arctan(tanx/√2）】（x是0~π/2）=√2π/4

Answer 2

The answer is π/(2√2)

∫<0,π/2> dx/(1+cos²x),用万能代换
令u=tan(x/2),dx=2/(1+u²) du,cosx=(1-u²)/(1+u²)
当x=0,u=0,当x=π/2,u=1
∴原式=∫<0,1> 1/[1+(1-u²)²/(1+u²)²]·2/(1+u²) du
=∫<0,1> (1+u²)/(1+u^4) du
=∫<0,1> {1/[2(u²+√2u+1)]+1/[2(u²-√2u+1)]}du
=(1/2)∫<0,1>du/(u²+√2u+1)+(1/2)∫<0,1>du/(u²-√2u-1)
=(1/2)∫<0,1>du/[(u+1/√2)²+1/2]+(1/2)∫<0,1>du/[(u-1/√2)²+1/2]
=(1/2)√2arctan[√2(u+1/√2)]<0,1>+(1/2)√2arctan[√2(u-1/√2)]<0,1>
=π/(8√2)+3π/(8√2)
=π/(2√2)
≈1.11072