Question

不定积分难题 求高手指点 万分感谢！！！！

∫dx/{cos²(2x)+2sin²(2x)}

Answer 1

令tanx=t,则x=arctant,dx=dt/(1+t^2),cos^2x=1/(1+t^2),sin^2x=1-cos^2t=t^2/(1+t^2)
∫dx/{cos²(2x)+2sin²(2x)}
=∫dx/[1+sin²(2x)]
=∫dx/(1+4sin²xcos²x)
=∫[dt/(1+t^2)]/[(1+4t^2/(1+t^2)^2]
=∫(1+t^2)/(t^4+6t^2+1)dt
=∫(1+t^2)/{[t^2+（√2+1)^2][t^2+（√2-1)^2]}dt
=[(√2+1)/（2√2)]∫dt/[t^2+（√2+1)^2]+[(√2-1)/（2√2)]∫dt/[t^2+（√2-1)^2]
=(√2/4){arctan[t/(√2+1)]+arctan[t/(√2-1)]}
=(√2/4){arctan[(√2+1)tanx]+arctan[(√2-1)tanx]}

Answer 2

分子分母同除以(cos(2x))^2，得：
∫dx/[cos²(2x)+2sin²(2x)]
＝∫(sec(2x))^2dx/[1＋2(tan(2x))^2]
＝1/2×∫1/[1＋2(tan(2x))^2] dtan(2x)
＝1/4×∫1/[1/2＋(tan(2x))^2] dtan(2x)
＝√2/4×arctan[√2×tan(2x)]＋C