求積分 1/4-cos^2x 和 1/4+cos^2x

應該是1/[4-(cosx)²] or 1/[4+(cosx)²]的積分吧!
1.∫ 1/[4-(cosx)²] dx
=∫ 2/[8-(1+cos2x)] dx=∫ 2/[7-cos(2x)] dx [令x=arctan(t) ]
=∫ 2/[7-(1-t²)/(1+t²)] *1/(1+t²) dt
=∫ 2/(6+8t²)dt
=∫ 1/[3+(2t)²] dt= (1/√3) arctan(2t/√3)+c
=(√3 /3)arctan(2tanx/√3) +c
2.∫1/[4+(cosx)²]dx
=∫ 2/[8+1+cos(2x)]dx (令x=arctan(t) )
=∫ 2/[9+(1-t²)/(1+t²)]*1/(1+t²)dt
= ∫ 1/[5+(2t)²] dt
= (1/√5) arctan(2t/√5)+c
=(√5 /5)arctan(2tanx/√5)+c

Note: cos(2x)=2(cosx)²-1, so, 2(cosx)²=1+cos(2x)
and cos(2x)=[1-(tanx)²]/[1+(tanx)²],
so if x=arctant, then cos(2x)=(1-t²)/(1+t²)

2010-05-30 00:04:37 補充：
x=arctan(t), then tanx=t
OK!?

括號的力量是不可忽略的

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1/4-cos^2x? 1/(4-cos^2x)?