∫sinxcosx/(1+sin^4x)dx
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∫sinxcosx/(1+sin^4x)dx∫sinxcosx/(1+sin^4x)dx∫sinxcosx/(1+sin^4x)dx∫sinxcosx/(1+sin^4x)dx=∫sinx/(1+si
∫sinxcosx/(1+sin^4x)dx
∫sinxcosx/(1+sin^4x)dx
∫sinxcosx/(1+sin^4x)dx
∫sinxcosx/(1+sin^4x)dx
=∫sinx/(1+sin^4x)d(sinx)
=1/2*∫1/(1+(sin^2x)^2)d(sin^2x)
=1/2*arctan(sin^2x)+C
令sinx=t
化简得
=∫t/(1+t^2)dx
=[ln(1+t^2)]/2+c
即：
ln(1+sinx^2)/2+c
∫sinxcosx/(1+sin^4x)dx
计算不定积分1.∫(sinXcosx)/(1+sin^4 X)dx 2.∫dx/(X^2 (4-X^2)^1/2)
计算不定积分1.∫(sinXcosx)/(1+sin^4 X)dx 2.∫dx/(X^2 (4-X^2)^0.5
已知tanx=2,则(1)1/4sin方x-3sinxcosx/cos方x+2sinxcosx
tanx=2,则1/4sin²x-3sinxcosx/cos²x+2sinxcosx=多少
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