# Math Help - I got problem integrating

1. ## I got problem integrating

The Problem is y' = cos(y - t), find general solution

My thought is to convert them to separable form

y - t = u ---> $dy/dt - 1 = du/dt$ --> Plug them back to original F

$du/dt + 1 = \cos u$

$du / ( \cos u - 1) = dt$

I integrate 2 sides
and I got stuck on this one $\int du / (\cos u - 1)$, please help me solve this problem
or if you have different method to find the general solution for the problem please help me
Thank you

2. Dear whitepenguin,

You could integrate $\int{\frac{1}{Cosu-1}dx}$ by the substitution, $Tan\frac{u}{2}=p$

Hope this helps.

3. Hello

When took calculus 2 . I really didn't have to solve this kind of problem using this method. Can you tell me when we should use this method ?

I only remember that I've used some sort of sub using x = a sint or x = a tant, x = a sect but never use this method... I'm pretty sure that they left this off , coz I didn't miss any day of my calculus class.

Thank you

4. Dear whitepenguin,

When you have an integral where the denominator contains only Cos or Sin terms of the first deree we use the substitution, $t=tan\frac{x}{2}$

eg: $\int{\frac{dx}{3Cosx-6Sinx+10}}$

Also for your knowledge it is good to remember when you have an integral where the denominator contains Sin and Cos terms of the second deree we can use the substituton $t=tanx$

eg: $\int{\frac{dx}{4Sin^2x+3Cos^2x+22}}$

If you have further questions regarding this matter please don't hesitate to reply me.

5. Originally Posted by whitepenguin
The Problem is y' = cos(y - t), find general solution

My thought is to convert them to separable form

y - t = u ---> $dy/dt - 1 = du/dt$ --> Plug them back to original F

$du/dt + 1 = \cos u$

$du / ( \cos u - 1) = dt$

I integrate 2 sides
and I got stuck on this one $\int du / (\cos u - 1)$, please help me solve this problem
or if you have different method to find the general solution for the problem please help me
Thank you
I think it might be easier to use the identity

$
\cos u = 1 - 2 \sin ^2 \frac{u}{2}
$
.