Troubleshooting Initial Value Problems for Differential Equations

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Homework Statement


Solve the Given initial value problem

Homework Equations



y' = tan ( x ), y ( pi / 4 ) = 3

The Attempt at a Solution



Well I got the answer

y = sec^2 ( x ) + 5 / 2

The books answer is

y = 3 - ln (sqrt(2) cos(x) )

Where did I got wrong? Or is the book screwed up?
 
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you integrated wrong, i don't know how they got sqrt(2) but you need to change tan(x) to sin(x)/cos(x) and do a substitution to get the log
 
What do YOU believe is the integral of tan(x)??
 
I am no expert, but I get the books answer, based on y=- ln (cos x)+C, and C =3 - ln[(sqrt2) /2)] , which = 3+ ln (sqrt 2).

Or y=-ln (cos x) -ln ( sqrt 2 ) +3

y= - ln [ (cos x) sqrt 2] +3

Dont bet your life on it. Jim
 
sheesh... I differentiated instead of integrated. It's been a long summer. Thanks for your time guys, appreciate it.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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