积分号dx/（5+4cos2x） 怎么积

加个好评哦
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令t=tanx; x=arctant;

dx=1/(1+t^2)dt;

cos2x=(1-t^2)/(1+t^2)

1/（5+4cos2x）=(1+t^2)/（9+t^2）;

积分函数变为 (1+t^2)/（9+t^2）*1/(1+t^2)dt=dt/(9+t^2)

因1/(a^2+x^2)的原函数是1/a*arctan(x/a)

故题目原函数是 1/3arctan(t/3)=1/3arctan(tanx/3)
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设F'(X)=5+4cos2X则F(X)=5X+2sin2X,自己积吧,你没给边界

设F'(X)=5+4cos2X
则F(X)=5X+2x * sin2X

你可用F(X)导一下检验~~