integrate 5x cube minus sin3x
\[\int\limits ((5x)^3-\sin(3x))dx\] you mean that?
well if that's what you mean you can make 2 integrals out of it and just integrate the left part and then the right part of it. need more help?
Yes I do
well now we have \[\int\limits (5x)^3 dx - \int\limits \sin(3x) dx\] well, ... i could write it out now for you but maybe you can remember some rules about integrating and that will help you. i can calculate it for you but that would not help you much i guess.
\[5 \int\limits x^3 dx - \int\limits \sin(3x) dx\] now just think of what equation you have to build the derivate of to get x^3 and sin 3x
x raised to power four over 4
right
that's right so we have the left side with \[\frac{125x^4}{ 4 }\] (i forgot the 5^3 before) and what about the right side? what is the derivate of sinus?
am going to guess here
minus cos
sinus --> cosinus --> minus sinus --> minus cosinus --> sinus ( the loop begins anew) so you have sin(3x) that means before that you hade a minus cosinus. therefore what is the solution for the right side?
plus cosx
plus cos3x
good but there is something missing. what happens if you build the derivate of the equation you just wrote? you get:\[-3\sin(3x)\] so you need to divide it with 3 and the whole endresult is: \[\int\limits ((5x)^3-\sin(3x) ) dx = \frac{125x^4} {4} + \frac{\cos(3x)}{3}\]
why divide by 3
was 3x diffrentiated
actually we divided cos 3x by 3 because while taking differential we get multiplied by 3
simply differentiate cos 3x/3 you will get back -sin 3x
chain rule
indeed
so integrate cos3x and divide by diffrention of 3x
is that it
well yes, but i wouldn't call it like that. just: but it's fine :)
Thank you
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