Notice that the original integral e70 sin(80) d0 is once again present. In fact, putting together all our work so far, if we let e70 sin(80) de, we have I =e70 sin(80) – e70 cos(80) + 1 + C1. 8 [1 717 I = 7 This can be re-arranged to give us 1e70 sin(80) - e70 cos(80) – I + C1. I =

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