Question

求定积分，详细点，谢谢，如图

Answer 1

解：设2u=√5tanz，则du=√5sec²zdz/2
于是，原式=∫<0,arctan(2/√5)>√5secz*√5sec²zdz/2
=(5/2)∫<0,arctan(2/√5)>sec³zdz
=(5/2)∫<0,arctan(2/√5)>coszdz/(cos²z)²
=(5/2)∫<0,arctan(2/√5)>d(sinz)/(1-sin²z)²
=(5/2)∫<0,arctan(2/√5)>(1/4)[1/(1-sinz)²+1/(1-sinz)+1/(1+sinz)²+1/(1+sinz)]d(sinx)
=(5/8)[1/(1-sinz)-ln│1-sinz│-1/(1+sinz)+ln│1+sinz│]│<0,arctan(2/√5)>
=(5/8)[ln│(1+sinz)/(1-sinz)│+2sinz/(1-sin²z)]│<0,arctan(2/√5)>
=(5/8)[2ln│(1+sinz)/cosz│+2sinz/cos²z]│<0,arctan(2/√5)>
=(5/4)[ln│secz+tanz│+secz*tanz]│<0,arctan(2/√5)>
=(5/4)[ln(3/√5+2/√5)+(3/√5)(2/√5)] (tanz=2/√5,secz=3/√5)
=(5/8)ln5+3/2。

Answer 2

u>0,设u=2分之根号5*tanX，du=2分之根号5*（1/cosx^2）dX，原式=10*∫（1/cosX^3）dX=10*∫（1/cosX^4）dsinX=10*∫(（1/(1-sinX^2）^2)dsinX，然后将被积函数拆分式啦，亲，