∫cos2x/(1+sinxcosx) dx 求详解.
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∫cos2x/(1+sinxcosx)dx求详解.∫cos2x/(1+sinxcosx)dx求详解.∫cos2x/(1+sinxcosx)dx求详解.Letu=1+sin(x)cos(x)=1+(1/
∫cos2x/(1+sinxcosx) dx 求详解.
∫cos2x/(1+sinxcosx) dx 求详解.
∫cos2x/(1+sinxcosx) dx 求详解.
Let u = 1 + sin(x)cos(x) = 1 + (1/2)sin(2x)
and du = cos(2x) dx → dx = du/cos(2x)
So ∫ cos(2x)/(1+sin(x)cos(x)) dx
= ∫ 1/u du
= ln|u| + C
= ln| 1 + sin(x)cos(x) | + C
or = ln| sin(2x) + 2 | + C
原式=∫cos2x/(1+(1/2)sin2x) dx
=∫(1/(1+(1/2)sin2x)d(1+(1/2)sin2x)
换元=∫(1/t)dt=ln(t)+c=ln(1+(1/2)sin2x)+c
cxcxcxcxc
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