∫sinx^2cosx^2dx 怎么解答 ∫sinx^2dx=
sinx^2cosx^2\u7684\u4e0d\u5b9a\u79ef\u5206\u600e\u4e48\u6c42sinx^2cosx^2\u7684\u4e0d\u5b9a\u79ef\u5206\u662fx/8-(sin4x)/32+C\u3002
sinx^2cosx^2
=[(sin2x)/2]^2
=[(sin2x)^2]/4
=(1-cos4x)/8.
\uff08sinx^2cosx^2\uff09
=(1/8)[x-(sin4x)/4]+C
=x/8-(sin4x)/32+C
\u6240\u4ee5sinx^2cosx^2\u7684\u4e0d\u5b9a\u79ef\u5206\u662fx/8-(sin4x)/32+C\u3002
\u6269\u5c55\u8d44\u6599\uff1a
1\u3001\u5e38\u7528\u51e0\u79cd\u79ef\u5206\u516c\u5f0f\uff1a
(1\uff09\u222b0dx=c
(2\uff09\u222b1/xdx=ln|x|+c
(3\uff09\u222be^xdx=e^x+c
(4\uff09\u222ba^xdx=(a^x)/lna+c
(5\uff09\u222bx^udx=(x^(u+1))/(u+1)+c
(6\uff09\u222bsinxdx=-cosx+c
\u5229\u7528\u4e8c\u500d\u89d2\u516c\u5f0f\uff0c\u697c\u4e0b\u7528\u5230\u4e86\uff0c\u4f46\u7ed3\u679c\u4e0d\u5bf9\uff1a
\u222bsinx^2dx
=\u222b\uff081-cos2x)dx/2
=\u222bdx/2-(1/2)\u222bcos2xdx
=(x/2)-(1/4)\u222bcos2xd(2x)
=(x/2)-(1/4)sin2x+c.
=(1/4) * ∫4sinx^2cosx^2dx
=(1/4) * ∫(2sinxcosx)^2dx (根据正弦倍角公式)
=(1/4) * ∫(sin2x)^2dx （根据余弦倍角公式）
=(1/8) *∫(1-cos4x)dx
=(1/8) *x - (1/8) * ∫cos4xdx + C (C是不定积分任意常数)
=(1/8) *x - (1/32) * ∫cos4xd4x + C (C是不定积分任意常数)
=(1/8) *x - (1/32)sin4x + C (C是不定积分任意常数)
=x/8 - sin4x/32 + C (C是不定积分任意常数)
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