求下列不定积分 ∫1/[sin(2x)+2sin x]dx
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求下列不定积分 ∫1/[sin(2x)+2sin x]dx
![求下列不定积分 ∫1/[sin(2x)+2sin x]dx](/uploads/image/z/12109945-49-5.jpg?t=%E6%B1%82%E4%B8%8B%E5%88%97%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86+%E2%88%AB1%2F%5Bsin%282x%29%2B2sin+x%5Ddx)
∫1/[sin(2x)+2sin x]dx
=∫1/[2sinxcosx+2sin x]dx
=∫1/(2sinx*[cosx+1]) dx
=∫1/(sinx*[2cos^2(x/2)]) d(x/2)
=1/2∫1/sinx dtan(x/2)
=1/2∫[1+tan^2(x/2)]/[2tan(x/2)] dtan(x/2)
=1/4∫[1/tan(x/2) ]+tan(x/2) dtan(x/2)
=1/4 ln|tan(x/2)|+1/8tan^2(x/2) +C
=∫1/[2sinxcosx+2sin x]dx
=∫1/(2sinx*[cosx+1]) dx
=∫1/(sinx*[2cos^2(x/2)]) d(x/2)
=1/2∫1/sinx dtan(x/2)
=1/2∫[1+tan^2(x/2)]/[2tan(x/2)] dtan(x/2)
=1/4∫[1/tan(x/2) ]+tan(x/2) dtan(x/2)
=1/4 ln|tan(x/2)|+1/8tan^2(x/2) +C