Integrating sqrt(x) cos(sqrt(x)) dx

[Phys12]

Question: sqrt(x) cos(sqrt(x)) dx
My try:

Let dv = cos(√x) => v = 2√xsin(√x) and u = √x => du = dx/(2√x)

Using integration by parts, we get

∫√x cos(√x) dx = 2√x√x sin(√x) - ∫(2√xsin(√x) dx)/(2√x)
= 2x sin(√x) - ∫sin(√x) dx

= 2x sin(√x) + 2 cos(√x) √x

However, the answer given in the book is: http://www.wolframalpha.com/input/?i=integrate+sqrt(x)++cos(sqrt(x))

And a solution that I found says: http://www.slader.com/textbook/9780534465544-calculus-early-transcendentals/601/61-exercises/23/

Which one of us is correct? And if I am wrong, what am I doing wrong and how may I correct it?

 

[nrqed]

Question: sqrt(x) cos(sqrt(x)) dx
My try:

Let dv = cos(√x) => v = 2√xsin(√x)

This is not correct.

 

[Phys12]

This is not correct.

I differentiated sin(sqrt(x)) and figured it out that way, why does it not work?

 

[nrqed]

I differentiated sin(sqrt(x)) and figured it out that way, why does it not work?

I realize this is what you did but why did you not differentiate also the $\sqrt{x}$ in front?

 

[Phys12]

I realize this is what you did but why did you not differentiate also the $\sqrt{x}$ in front?

Inside the sin()? I did do that

 

[nrqed]

Inside the sin()? I did do that

No, the $\sqrt{x}$ in **front** of the sine.

 

[Phys12]

No, the $\sqrt{x}$ in **front** of the sine.

Well, this is what I did:

d(sin(sqrt(x)))/dx = cos(sqrt(x))/2(sqrt(x) => integral cos(sqrt(x)) = 2 sqrt(x) sin(sqrt(x))

 

[nrqed]

Well, this is what I did:

d(sin(sqrt(x)))/dx = cos(sqrt(x))/2(sqrt(x) => integral cos(sqrt(x)) = 2 sqrt(x) sin(sqrt(x))

You cannot move the $\sqrt{x}$ on the other side like that (in your last step).
What you showed is that
$$ \frac{ d}{dx} \sin(\sqrt{x}) = \frac{\cos(\sqrt{x})}{2 \sqrt{x}} $$

which means the following

$$ \int \frac{\cos(\sqrt{x})}{2 \sqrt{x}} = \sin(\sqrt{x}) +C $$ This is all that one can conclude from your calculation. So your v is incorrect

 

[Phys12]

You cannot move the $\sqrt{x}$ on the other side like that (in your last step).
What you showed is that
$$ \frac{ d}{dx} \sin(\sqrt{x}) = \frac{\cos(\sqrt{x})}{2 \sqrt{x}} $$

which means the following

$$ \int \frac{\cos(\sqrt{x})}{2 \sqrt{x}} = \sin(\sqrt{x}) +C $$ This is all that one can conclude from your calculation. So your v is incorrect

So I used integration by parts and got the correct answer, however, now, for the original question, I got this:

sqrt(x) [2 sqrt(x) sin(sqrt(x)) + 2 cos(sqrt(x))] - int (sin(sqrt(x)) dx) - int(cos(sqrt(x)) dx/sqrt(x))
and when I solve further, I get:

int (sin(sqrt(x)) dx) = -2 cos(x) x + 2 sin(x)

and int(cos(sqrt) dx/sqrt(x)) = 2 sin(sqrt(x))

and end up getting the wrong answer. :/ What am I doing wrong?