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Joined: 25/09/2014 - 00:04
Posts: 1



Where do i start??

Joined: 24/01/2015 - 05:29
Posts: 2
how to start?... hmmm... tricky disco

I would suggest reading the problem specification again, because I don't think your teacher assigned that.

What I suspect your question really was would have the sin(x) replaced with sin(x+5). That's doable.

So I'm going to assume you want something like this: Int[-sin(x+5)ln(cos(x+5))]dx.

First, let's do a u substitution with u=x+5. Then du=dx, and Int[-sin(u)ln(cos(u))]du.

Now let's do another, say, v substitution, with v=cos(u). Then dv=-sin(u)du, and Int[ln(v)]dv.

Now, that's in our table of integrals, and we see the result is v(ln(v)-1) + C

where C is the constant of integration. Of course, that's not our final answer, because we still

need to do back substitution to find the result in terms of x, which is how the problem is posed.

So, v=cos(u), meaning our intermediate answer (in terms of u) is: cos(u)(ln(cos(u))-1)+C.

And u=x+5, so our final answer would be: cos(x+5)(ln(cos(x+5))-1)+C.


Now, we have *an* answer. But is this the correct answer? Let's take the derivative to check.

Let y(x)=cos(x+5)(ln(cos(x+5))-1)+C

     y'(x)=-sin(x+5)(ln(cos(x+5))-1)+cos(x+5)(-sin(x+5)/(cos(x+5)) by the product rule

             =-sin(x+5)ln(cos(x+5))+sin(x+5)-sin(x+5)                             expansion/simplifying

             =-sin(x+5)ln(cos(x+5))                                                              final form, and it matches.


I know this is not your quesiton, but.. this is how you do one of these, however your problem is harder,

and I highly doubt it is the one you are trying to solve. If it is, try wolfram alpha or another CAS for it.

Sorry if this wasn't what you were looking for... (I think it is though :)