escobar147
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limits from τ=0 to t
∫τsinωτ dτ
answer: [-τ/ω + 1/ω^2(sinωτ] limits τ = 0 to t
= -t/ω(cosωt) + 1/ω^2(sinωt)Could someone please explain the steps? is this an integration by parts method??
when i attempt integration by parts, with u = τ & dv = sin(ωτ) dτ
I get the following answer:
τ(ωcosωτ) - sinωτ
∫τsinωτ dτ
answer: [-τ/ω + 1/ω^2(sinωτ] limits τ = 0 to t
= -t/ω(cosωt) + 1/ω^2(sinωt)Could someone please explain the steps? is this an integration by parts method??
when i attempt integration by parts, with u = τ & dv = sin(ωτ) dτ
I get the following answer:
τ(ωcosωτ) - sinωτ
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