∫ (1 + cos ^ 2 + tan ^2 ) sec^2 dx
∫ (1 + cos ^ 2 + tan ^2 ) sec^2 dx
∫sec^4x dx ∫sec^2x tan^2x dx
∫tan^2xdx=∫(sec^2x-1)dx
∫secx/sec^2x-1 dx
∫ sec^2 x dx
∫(1+cos^3 x)sec^2 x dx 求不定积分~
∫cosx cos3x dx ∫tan^3t sect dt ∫(sec^2x)/4+tan^2 dx
=∫dx/(2cos^2(x/2)+1)=∫sec^(x/2)dx/(2+sec^2(x/2))怎么出来的?
三角函数 sec(2a)=sec^a/1-tan^a
求积分∫(sec^2x/2+tan^2x)dx
(1) 证明1+sec a+tan a/1+sec a -tan a=1+sin a /cosa (2)sinx+cos
∫(sec^2x+sinx)dx