求不定积分 ∫ sin^4/cos^2 dx
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求不定积分 ∫ sin^4/cos^2 dx
![求不定积分 ∫ sin^4/cos^2 dx](/uploads/image/z/1246593-57-3.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86+%E2%88%AB+sin%5E4%2Fcos%5E2+dx)
原式=∫[1-(cosx)^2]^2/(cosx)^2dx
=∫[(secx)^2-2+(cosx)^2]dx
=∫(secx)^2dx-2x+(1/2)∫(1+cos2x)dx
=tanx-2x+x/2+(sin2x)/4+C
=tanx-3x/2+(sin2x)/4+C.
=∫[(secx)^2-2+(cosx)^2]dx
=∫(secx)^2dx-2x+(1/2)∫(1+cos2x)dx
=tanx-2x+x/2+(sin2x)/4+C
=tanx-3x/2+(sin2x)/4+C.