Question

∫1/(sin^2x+cos^4x)dx求原函数

Answer 1

∫dx/[(sinx)^2+(cosx)^4]

分子分母同时除(cosx)^4
=∫(secx)^4/[(secx)^2+1] dx
=∫(secx)^2/[(secx)^2+1] dtanx
=∫ { 1 - 1/[(secx)^2+1] } dtanx
=∫ { 1 - 1/[(tanx)^2+2] } dtanx
=tanx -∫ dtanx/[(tanx)^2+2]
=tanx -(1/2) ∫ dtanx/[1+ ((tanx)/√2)^2]
=tanx -(√2/2) ∫ d(tanx/√2)/[1+ ((tanx)/√2)^2]
=tanx -(√2/2)arctan[(tanx)/√2] + C

Answer 2

同学你好
分解方式如下：
∫dx/[(sinx)^2+(cosx)^4]
分子分母同时除(cosx)^4
=∫(secx)^4/[(secx)^2+1] dx
=∫(secx)^2/[(secx)^2+1] dtanx
=∫ { 1 - 1/[(secx)^2+1] } dtanx
=∫ { 1 - 1/[(tanx)^2+2] } dtanx
=tanx -∫ dtanx/[(tanx)^2+2]
=tanx -(1/2) ∫ dtanx/[1+ ((tanx)/√2)^2]
=tanx -(√2/2) ∫ d(tanx/√2)/[1+ ((tanx)/√2)^2]
=tanx -(√2/2)arctan[(tanx)/√2] + C

Answer 3

∫(sin4x+cos4x)dx=1/4*∫(sin4x+cos4x)d4x=1/4*(-cos4x+sin4x)+C