Question

求反三角函数定积分问题，见图片。谢谢！

Answer 1

设 t = arctan(1/x)，当 x = 0 时，t = π/2；当 x = 1 时，t = π/4。
则 x =1/(tant)，dx = -(sect)^2 *dt/(tant)^2 = - dt/(sint)^2
那么，原积分可以变换为：
∫t * (-dt)/(sint)^2 注：积分限变成：π/2 ~ π/4
=∫t*[-(cscx)^2]*dt
=t*cot(t) - ∫cot(t)*dt
=t*cot(t) - ∫cost *dt/sint
=t*cot(t) - ∫d(sint)/sint
=t*cot(t) - ln|sint|
=[π/4*cot(π/4) - π/2 * cot(π/2)] - [ln|sin(π/4)| - ln|sin(π/2)|
=[π/4* 1 - π/2 * 0] - [ln(1/√2) - 0]
=π/4 + 1/2*ln2