Question

定积分求解 ，如图

Answer 1

约定:∫[a,b]表示[a,b]上的定积分

原式=∫[0,π/2](sin²t/cost)d(sint) (设x=sint)
=∫[0,π/2]((sin²t/cost))costdt
=∫[0,π/2](sin²t)dt
=(1/2)∫[0,π/2](1-cos2t)dt
=((t/2)-(1/4)sin2t)|[0,π/2]
=(π/4)-0
=π/4

希望能帮到你！

Answer 2

Answer 3

Answer 4

x=1是被积函数的瑕点
令x=sint,则dx=costdt,√(1-x²)=cost
当x=0时,t=0.x=1时,t=π/2
原式=∫[0,π/2]sin²tcostdt/cost
=∫[0,π/2]sin²tdt
=∫[0,π/2](1-cos2t)/2*dt
=∫[0,π/2]dt/2-∫[0,π/2]cos2td2t/4
=t/2|[0,π/2]-sin2t/4|[0,π/2]
=π/4