Finding $ \int_{0}^{1} 2x\sin(\frac{1}{x}) - \cos(\frac{1}{x})\ dx. $ How do we solve the following integral ?
$  \int_{0}^{1} 2x\sin(\frac{1}{x}) - \cos(\frac{1}{x})\ dx.  $
I tried to proceed by integration by parts but got stuck. 
 A: I think I have a guess for this one. But it is only a guess. This looks like an expansion of product rule. In particular, notice that 
$$
\frac{d}{dx}\left(x^2\sin(1/x)\right) = 2x\sin(1/x) + x^2\cos(1/x)(-1/x^2) = 2x\sin(1/x) - \cos(1/x)
$$
So, we get
$$
\int_0^1 2x\sin(1/x) - \cos(1/x) dx = \left[x^2\sin(1/x)\right]_0^1 = \sin(1)
$$
I realize this isn't very rigorous. I think we need to consider a limit for the $0$ side
$$
\lim_{x\to 0}x^2\sin(1/x) = 0
$$
which follows from by the squeeze theorem.
A: $$\int 2x\sin(1/x)dx -\int \cos(1/x)dx$$
In fact, you can only do integration by parts only once, 
$$\int 2x\sin(1/x)dx=\int \cos(1/x)dx+x^2\sin(1/x)$$
By Letting $$f'(x)=2x \implies f(x)=x^2$$
$$g(x)=\sin(1/x) \implies g'(x)=-x^2\cos(1/x)$$
$$\therefore \int 2x\sin(1/x)dx -\int \cos(1/x)dx=\int \cos(1/x)dx+x^2\sin(1/x) - \int \cos(1/x)dx$$ $$=x^2\sin(1/x)$$
Now, you need only find the definite integral.
$$\sin(1)-\lim_{x \to 0} x^2\sin(1/x)$$
Using the squeeze theorem, 
$$-1<\sin(1/x)<1 \implies 0<\lim_{x \to 0} x^2\sin(1/x)<0$$
So, the limit is equal to zero.
$$\int_{0}^{1} 2x\sin(1/x) - \cos(1/x)\ dx =\sin(1)$$
A: You can do this using int by parts only once.
First I split it into a sum of integrals:
$$\int_0^1 2x\sin(1/x)dx - \int_0^1\cos(1/x)dx$$
Then integrate the first integral only:
I did this with integration by parts $u = sin(1/x)$ and $dv = 2x$. This gives:
$$\left[uv-\int vdu\right] - \int cos(1/x)$$
$$\left[x^2\sin(1/x)|_0^1 - \int_0^1-\frac{x^2}{x^2}\cos(1/x)dx\right] - \int_0^1 \cos(1/x) dx$$
$$x^2\sin(1/x)|_0^1 + \left[\int_0^1\cos(1/x)dx - \int_0^1 \cos(1/x) dx\right]$$
Then the $\int cos(1/x)dx$ cancels out. So evaluate $x^2\sin(1/x)]_0^1$. At $0$ use squeeze theorem as @Alec said.
The final answer is $1\sin(1)$
A: Set $u'=2x$ and $v=\sin(1/x)$ and apply $\int u' v = u v - \int u v'$. Now use the integration bounds.
A: Before evaluating the definite integral, I will evaluate the indefinite integral.
$$~~\int 2x\sin(1/x) - \cos(1/x)\,dx\\
 = \int 2x\sin(1/x) - \int \cos(1/x)\,dx$$
The first integral is evaluated as such:
$$\int 2x\sin(1/x) = 2\int x\sin(1/x)$$
$$ u = x, ~du=dx, ~dw=\sin(1/x), ~w = x \sin(1/x)-Ci(1/x)$$
Where $Ci(x)$ is the cosine integral, $Ci(x) = -\int_x^{\infty}\frac{\cos t}{t}dt$ and $\frac{d}{dx}Ci(x) = \frac{\cos(x)}{x} $
This cos integral is probably the first place you got stuck, it is found by integration by parts:
$$\int \sin(1/x)~dx \\
a = \sin(1/x),~ da = -\cos(1/x)/x^2~dx,~ db = dx,~ b = x \\
\int a~db = ab - \int b ~da \\
\int \sin(1/x) = x\sin(1/x)- \int \cos(1/x)/x  = x\sin(1/x) - Ci(1/x)$$
Anyway,
$$\int x\sin(1/x) = x^2 \sin(1/x)-xCi(1/x) - \int x\sin(1/x)-Ci(1/x)~dx$$
Next, By a similar method,
$$\int \cos(1/x) = \frac{sin(1/x)}{x^2}$$
Therefore,
$$\int 2x\sin(1/x) - \cos(1/x)\,dx = \\x^2 \sin(1/x)-xCi(1/x) - \int x\sin(1/x)-Ci(1/x)~dx - \frac{sin(1/x)}{x^2}$$
And I'll leave you to evaluate the rest.
Here are some referances:


*

*Cosine Integral

*Sine Integral
