Math Help - Calculus quiz tomorrow..help with this one review problem?

1. Calculus quiz tomorrow..help with this one review problem?

Solve initial value problem using sep. of variables

y(prime)=cos^2(x) y(0)=pi/4

I ended up using integration by parts and got to here:

y=cosx * sinx - x + C

and C= pi/4

but the book says the answer is:

y=.5(x+.5sin(2x))+pi/4

what did I do wrong?

2. Originally Posted by woohoo
Solve initial value problem using sep. of variables

y(prime)=cos^2(x) y(0)=pi/4

I ended up using integration by parts and got to here:

y=cosx * sinx - x + C

and C= pi/4

but the book says the answer is:

y=.5(x+.5sin(2x))+pi/4

what did I do wrong?
Use the identity $\cos^2{x} = \frac{1}{2}(1 + \cos{2x})$.

So $y = \int{\cos^2{x}\,dx} = \frac{1}{2}\int{(1 + \cos{2x})\, dx}$

$= \frac{1}{2}(x + \frac{1}{2}\sin{x}) + C$

Since $y(0) = \frac{\pi}{4}$,

$\frac{\pi}{4} = \frac{1}{2}(0 + \frac{1}{2}\sin{0}) + C$

$C = \frac{\pi}{4}$

So $y = \frac{1}{2}(x + \frac{1}{2}\sin{x}) + \frac{\pi}{4}$.

3. Okay I understand the book's answer now.

But I don't get why my method turned out incorrect.

I got to where y= integral (cos^2(x) dx)=integral (cosx * cosx dx)

using separation by parts:

y=cosxsinx-integral(x)=cosxsinx-x+C

where did I go wrong?

4. Nevermind.

I figure out where I went wrong.

Thanks!

5. Originally Posted by woohoo
Okay I understand the book's answer now.

But I don't get why my method turned out incorrect.

I got to where y= integral (cos^2(x) dx)=integral (cosx * cosx dx)

using separation by parts:

y=cosxsinx-integral(x)=cosxsinx-x+C

where did I go wrong?
Are you trying to do separation of variables or integration by parts? They're two different methods.

You only use separation of variables if you have a Separable linear DE, which you don't, so I guess you're trying to do integration by parts...

Integration by parts is as follows...

$\int{u\,dv} = uv - \int{v\, du}$.

So if you're integrating $\cos{x}\times\cos{x}$, then $u = \cos{x}$ and $dv = \cos{x}$. So what are $du$ and $v$?

In this case you should end up with $\int{\sin^2{x}\,dx}$... not fun to work with as it's just as bad as $cos^2{x}$. Just use the trig identity.