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What is $\int 2x \cos(x^2 + 1) dx$? ok, so this is driving me crazy. I use integration by parts so: $$\int f(x)g'(x)dx = f(x)g(x)-\int f'(x)g(x)dx$$ $$f(x)= 2x \qquad g(x)=\cos(x^2+1)$$ $$f'(x)=2 \qquad g'(x)=-2x\sin(x^2+1)$$ Now I apply the formula( as only one side of the equation is enough I will do that on the right hand site of it i.e: $f(x)g(x)-\int f'(x)g(x)dx$ so: $$2x * \cos(x^2+1) - \int 2 * \cos(x^2+1)dx$$ now I find myself pretty much on the same spot I was before. What am I doing wrong? I am just following step by step the formula...

Then, not sure if related or not, but watching videos on youtube about that I found the following $\int e^{2x} * \sin(x) dx$,

the guy uses the formula $uv-\int vdu$

So, he takes $u=e^{2x}$ and $u'=2e^{2x}$

I agree with him so far but then he goes and $v = -\cos x$ and $du = \sin x dx$

why is he doing that?? I would put $v=\sin(x)$ and $v'= \cos(x)$

I see he is differentiating my $v$ for some reason , or trying top, as the derivative of

$\sin(x)$ is $\cos(x)$ not $-\cos(x)$....why is all that about? why isnt he following the formula?

people seem to be happy in their comments with what he does, so it must make sense

somehow.. he ends up writing: $-e^{2x}*\cos(x) + 2 \int e^x \cos (x) dx$

I would have writtn instead, $e^{2x}*\sin(x)-\int\sin(x)*2e^{2x}$...so totally different... again

why?? I am just sticking to the formula....

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4  
just by looking at it... you should have substituted u = $x^2+1$ then you will have $du= 2x .dx $ –  Dreamer78692 Feb 18 '13 at 7:02
    
Note that $2x = \frac{d}{dx} (x^2+1)$. –  copper.hat Feb 18 '13 at 7:03
3  
The integration by parts formula is about $\int f(x)g'(x)\,dx$. You had let $f(x)=2x$ and $g(x)=\cos(x^2+1)$. But if you will use integration by parts, it is $g'(x)$ that you must choose, and then you find $g(x)$. (And as pointed out by others, the whole problem yields to a natural substitution.) –  André Nicolas Feb 18 '13 at 7:09
    
How urgent is this? –  Mariano Suárez-Alvarez Feb 18 '13 at 7:21
3  
Homework is always urgent... –  copper.hat Feb 18 '13 at 7:23

5 Answers 5

You don't need to use integration by parts. Consider $$z=x^2+1,$$ then, $$dz=2xdx.$$ Thus, $$\int 2x\cos(x^2+1)dx=\int \cos(z)dz=\sin(z)=\sin(x^2+1).$$

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I wouldn't solve this one by Integration with parts i would use the substitution rule so $$\int 2 x \cos(1+x^2)\, \mathrm{d}x=\int \cos(u) \, \mathrm{d}x = \sin(u)+ c = \sin(x^2+1)+c$$ where $c$ is an arbitrary constant, an $u=x^2+1$

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Just for something slightly different...

\begin{eqnarray} \int 2x \cos(x^2+1) dx &=& \int 2x \sum_n (-1)^n \frac{(1+x^2)^{2n}}{(2n)!} dx \\ &=& \sum_n (-1)^n \int 2x \frac{(1+x^2)^{2n}}{(2n)!} dx \\ &=& \sum_n (-1)^n \frac{(1+x^2)^{2n+1}}{(2n+1)(2n)!} \\ &=& \sin (1+x^2) \end{eqnarray}

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Of course, this is still using the same substitution... –  copper.hat Feb 18 '13 at 7:25
    
you forgot the arbitrary constant, just saying –  Dominic Michaelis Feb 18 '13 at 7:49
    
I didn't. I just didn't put it in :-). –  copper.hat Feb 18 '13 at 7:51

You shouldn't integrate by parts to solve this integral $\int 2x\cos(x^2+1)dx$. This is the integral of a product where $2x$ is exactly the extra term when you take the derivative of $f(x^2+1)$ for some $f$, so you in fact $\sin(x^2+1) + C$ is an antiderivative: $\sin$ becomes $\cos$ and the extra term is then just what you need...

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For the first one, note that $\frac{d}{dx}\sin(x^{2}+1)=2x\cos(x^2+1)$, so there's no need for integration by parts. But if you must integrate by parts, let $f(x) = 1, g(x)=\sin(x^{2}+1)$, (i.e. $g'(x)=2x\cos(x^2+1)$).

For the second part of your question, the formula for integration by parts is $\int udv = uv-\int vdu$. So either $e^{2x}$ or $\sin x$ must be the $dv$ term. In his approach, he went with $dv=\sin x$.

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f(x)= 1? and g(x)=sin(x^2+1)? f(x)=2x and g(x)=cos(x^2+1)...what I am missing?? on the second part of the answer..the dv term? dv term is not v differentiated? so cos(x)?? –  Maximilian1988 Feb 18 '13 at 7:23
    
In doing integration by parts you have to have a product of a function and a derivative of some other function, and you have to pick smartly. So if you pick $f(x)=2x$, you're left with $g'(x)=cos(x^{2}+1)$, and you'll find it hard to find an antiderivative for that. In the second part, the dv term is v differentiated, but in the integral you have a product of two functions, namely $e^{2x}$ and $sinx$. As said, you can't choose both of them as being u and/or v. So then you are left with either 1) $du = e^{2x}$, $v=sinx,$ or 2) $u=e^{2x}$, $dv = sinx$. In the end, both work. –  Ryker Feb 18 '13 at 16:58

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