Question

∫1/[2+tan^2(x)]dx=多少

Answer 1

∫ 1/(2 + tan²x) dx,u = tanx,du = sec²x dx

= ∫ 1/(2 + u²) · du/sec²x

= ∫ 1/(2 + u²) · 1/(1 + u²) · du

= ∫ [(u² + 2) - (u² + 1)]/[(u² + 2)(u² + 1)] · du

= ∫ du/(u² + 1) - ∫ du/(u² + 2)

= arctan(u) - (1/√2)arctan(u/√2) + C

= arctan(tanx) - (1/√2)arctan((tanx)/√2) + C

= x - (1/√2)arctan((tanx)/√2) + C