calculus
posted by Anna .
(SIN^8X5 SIN^5X+SINX)COSX DX
I need the answer to this, because I have tried so many times and cannot figure it out.

I infer from the presence of DX that you want
∫(sin^8(x)  5sin^5(x) + sin(x)) cos(x) dx
if you let u = sin(x), then du = cos(x) dx you have
∫(u^8  5u^5 + u) du
that's dead easy, using thge power rule:
1/9 u^9  1/6 u^6 + 1/2 u^2 + C
now change all the u's back to sin(x) and you're done
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