indefinite integral cos(t^1/2)/(t^1/2)
\[\int\limits_{}^{}\frac{ \cos \sqrt{t} }{ \sqrt{t} }\]
try \(u=\sqrt{t}\)
@satellite73 always beats me to the punch lol :D
i dont understand how u subtitution really works. can someone explain what do i do after finding u?
@satellite73 @jagr2713
ok lets say you have 5(2x+3)=3(4x+8) you just multiply 5x2 and 5x3 and 3x4 and 3x8 so you will get 10x+15=12x+24 then you just solve for x
i still dont understand. once i find out u = sqrt(t) then what do i do?
@jagr2713
oh you mean substitution of your problem. i dont know how to do that sorry
rewrite the integral in terms of u and du. So what is du?
that is give me \[D[\sqrt{t}]\]
du=1/2sqrt(t) @sylbot
so?? just plug it in u = \(\sqrt t\rightarrow du=\dfrac{1}{2\sqrt t}dt\) your integral turns to \[\int 2cosu du= 2\int cos u du\] It's quite easy now, right?
the derivative of \[\sin(\sqrt{t})\] is \[\cos(\sqrt{t})\times \frac{1}{2\sqrt{t}}\] by the chain rule that would be your answer, except you are off by a factor of two
Looks like you've got plenty of quality help already, I just wanted throw in my two cents. Personally I like to rewrite the integral, separate things a little bit, so it will be easier to see where everything is getting plugged in.\[\Large\rm \int\limits \frac{\cos \sqrt t}{\sqrt t}dt=\int\limits \cos\color{royalblue}{\sqrt t}\left(\color{orangered}{\frac{1}{\sqrt t}dt}\right)\]Then you're making the substitution:\[\Large\rm \color{royalblue}{u=\sqrt t}\]Taking your derivative gives,\[\Large\rm du=\frac{1}{2\sqrt t}dt\qquad\to\qquad \color{orangered}{2du=\frac{1}{\sqrt t}dt}\]So then plugging in our pieces,\[\Large\rm \int\limits\limits \cos\color{royalblue}{\sqrt t}\left(\color{orangered}{\frac{1}{\sqrt t}dt}\right)=\int\limits\limits \cos\color{royalblue}{u}\left(\color{orangered}{2du}\right)\]
Thank you actually the last one helped me out the most understanding!
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