∫sixcosx/1+sin^4x dx请你详解
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∫sixcosx/1+sin^4xdx请你详解∫sixcosx/1+sin^4xdx请你详解∫sixcosx/1+sin^4xdx请你详解∫sinxcosx/(1+sin^4x)dx=∫sinx/(1
∫sixcosx/1+sin^4x dx请你详解
∫sixcosx/1+sin^4x dx
请你详解
∫sixcosx/1+sin^4x dx请你详解
∫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
∫sixcosx/1+sin^4x dx请你详解
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