∫[0→π/2]
(cosx)^5sinx
dx
=-∫[0→π/2]
(cosx)^5
d(cosx)
=-(1/6)(cosx)^6
|[0→π/2]
=1/6
∫
(x³-x)/(x+1)
dx
=∫
(x³+x²-x²-x)/(x+1)
dx
=∫
(x³+x²)/(x+1)
dx
-
∫
(x²+x)/(x+1)
dx
=∫
x²
dx
-
∫
x
dx
=(1/3)x³
-
(1/2)x²
+
C
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