Question

How to integrate 4 sin 2t dt?

Answer 1

\[\int\limits_{}^{}4\sin2t dt\]
let u = 2t, therefore
du = 2dt
dt = du/2
\[\int\limits_{}^{}4sinu*\frac{ du }{ 2 }= 2\int\limits_{}^{}sinu\]

the integral of sinu is -cosu, meaning our answer is:

\[-2cosu + C = -2\cos(2t) + C\]
At the end I just replaced u with what I originally said it was.

Answer 2

Ok I got it. Thanks Psymon. :)

Answer 3

Okay, awesome.

Answer 4

How about integrate -y(1+2t^2) dt?

Answer 5

-y and t in the same equation?

Answer 6

Ops sorry.Wrong typing. It's -(1+2t^2) divide t dt.

Answer 7

So:
\[\frac{ -(1+2t^{2}) }{ t }\]
?

Answer 8

Yes it is.

Answer 9

Gotcha. Well, split it up into two fractions:
\[\frac{ -(1+2t^{2}) }{ t }= -\frac{ 1 }{ t }-\frac{ 2t^{2} }{ t } \implies -\int\limits_{}^{}\frac{ 1 }{ t }-2\int\limits_{}^{}t\]As long as you know how to integrate those now youre good.

Answer 10

Ok so I will get - In t - 2t^2 /2.Right?