Question

求高数问题详细过程指导

Answer 1

令x=atant，则dx=asec^2tdt

原式=∫a^2tan^2t/asect*asec^2tdt
=a^2*∫tan^2tsectdt
=a^2*∫tantd(sect)
=a^2*tantsect-a^2*∫sec^3tdt
=a^2*tantsect-a^2*∫sect(tan^2t+1)dt
=a^2*tantsect-a^2*∫tan^2tsectdt-a^2*∫sectdt
=a^2*tantsect-a^2*∫tan^2tsectdt-a^2*ln|sect+tant|
所以a^2*∫tan^2tsectdt=(a^2/2)*[tantsect-ln|sect+tant|]+C
原式=(x/2)*√(a^2+x^2)-(a^2/2)*ln|x+√(a^2+x^2)|+C，其中C是任意常数