sint=x,cos2t怎么用x表示出来(x^2<1)?
\u222b1/(1+(1-x^2)^(1/2))dx\u9898\u4e2d\uff0c\u7528x=sint\u4ee3\u5165\u7684\uff0c\u505a\u5230\u4e3at-tan(t/2)+c\uff0c\u95ee\u7528x\u8868\u793a tan\uff08t/2\uff09?= t - tan(t/2) + C= arcsinx - sin(t/2)/cos(t/2) + C
= arcsinx - [2sin(t/2)cos(t/2)]/[2cos²(t/2)] + C
= arcsinx - sint/(1 + cost) + C
= arcsinx - x/(1 + \u221a(1 - x²)) + C
\u2235x = sint\uff0c\u2234cost = \u221a(1 - x²)
\u222b(1-x^2)^0.5dx=\u222b\u221a(1-x^2)dx
\u8bbex=sint\uff0cdx=costdt\uff0ct=arcsinx
\u539f\u5f0f=\u222b\u221a[1-(sint)^2]*costdt
=\u222b(cost)^2dt
=\u222b(1+cos2t)/2*dt
=\u222b(1/2)dt+\u222b(1/2)cos2tdt
=t/2+\u222b(1/4)cos2td(2t)
=t/2+(1/4)sin2t+C
=t/2+(1/2)sintcost+C
=t/2+(1/2)sint*\u221a[1-(sint)^2]+C
=(1/2)arcsinx+(1/2)x\u221a(1-x^2)+C
所以 (sint)^2=x^2
因为 (sint)^2+(cost)^2=1
所以 (cost)^2=1-x^2
所以 cos2t=(cost)^2-(sint)^2
=(1-x^2)-x^2
=1-2x^2.
绛旓細鍥犱负 (sint)^2+(cost)^2=1 鎵浠 (cost)^2=1-x^2 鎵浠 cos2t=(cost)^2-(sint)^2 =(1-x^2)-x^2 =1-2x^2.
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