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Question Number 34382 by rashedul last updated on 05/May/18
how do i prove  ∫_o ^t cos 2(ωt+α)dt=0
$$\mathrm{how}\:\mathrm{do}\:\mathrm{i}\:\mathrm{prove} \\ $$$$\int_{\mathrm{o}} ^{\mathrm{t}} \mathrm{cos}\:\mathrm{2}\left(\omega\mathrm{t}+\alpha\right)\mathrm{dt}=\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 05/May/18
=((∣sin(2wt+2α)∣_0 ^T )/(2w))  =((sin(2wT+2α)−sin(2α))/(2w))  w=((2Π)/T) so    =((sin(4Π+2α)−sin2α)/(2ω))  =((sin2α−sin2α)/(2w))  =0  i think upper limig of intregal is T where T  T=2Π/w
$$=\frac{\mid{sin}\left(\mathrm{2}{wt}+\mathrm{2}\alpha\right)\underset{\mathrm{0}} {\overset{{T}} {\mid}}}{\mathrm{2}{w}} \\ $$$$=\frac{{sin}\left(\mathrm{2}{wT}+\mathrm{2}\alpha\right)−{sin}\left(\mathrm{2}\alpha\right)}{\mathrm{2}{w}} \\ $$$${w}=\frac{\mathrm{2}\Pi}{{T}}\:{so}\:\: \\ $$$$=\frac{{sin}\left(\mathrm{4}\Pi+\mathrm{2}\alpha\right)−{sin}\mathrm{2}\alpha}{\mathrm{2}\omega} \\ $$$$=\frac{{sin}\mathrm{2}\alpha−{sin}\mathrm{2}\alpha}{\mathrm{2}{w}} \\ $$$$=\mathrm{0} \\ $$$${i}\:{think}\:{upper}\:{limig}\:{of}\:{intregal}\:{is}\:{T}\:{where}\:{T} \\ $$$${T}=\mathrm{2}\Pi/{w} \\ $$

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