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() | sin3x dx â |
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Answer» Note thatsin2x=1−cos2xsin2x=1−cos2x. Then: ∫sinx(1−cos²x)dx =∫sinx−sinxcos²xdx= =−cosx−∫sinxcos²xdx= Letu=cosx⟺du=−sinxdx, so: =−cosx+∫u²du= =−cosx+1/3u³+C=1/3cos³x−cosx find the value integration sin3xdx |
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