Calculate the derivative of underintegral function and find the integral I have the integral:
$$ \Phi(\alpha)=\int\limits_{1}^{2\alpha}\frac{\cos(2\alpha x^3)}{x}\,\mathrm{d}x \tag{1}$$
Have to find $\Phi'(\alpha)$ and calculate the integral then.
do I understand correctly that I should find $\frac{d\Phi}{d\alpha}$ first and get:
$$-\frac{1}{x}\sin(2\alpha x^3)\cdot2x^3$$
and then plug the answer in into the original integral and get:
$$-2\int\limits_{1}^{2\alpha}x^2\sin(2\alpha x^3)\,\mathrm{d} \tag{2}x$$
Now $(2)$ looks like this need to be taken by parts, but even with this knowledge I have no idea what to do regards $\alpha$
 A: You did the first part great:
$$I(a)=\int_{1}^{2a}\frac{\cos(2a x^3)}{x}dx\Rightarrow I'(a)=-2\int_1^{2a}x^2 \sin(2a x^3)dx$$
Now look that we have $x^3$ that buggs us inside the sine function, but also an $x^2$ that is almost the derivative of $x^3$. So indeed we can make the substitution of $x^3=t$ that will help us alot:
$$I'(a)=-\frac23\int_1^{(2a)^3} \sin(2at)dt=\frac23\frac{ \cos (2at)}{2a}\bigg|_1^{(2a)^3}=\frac{\cos((2a)^4)-\cos(2a)}{3a}$$ 
Now in order to find the original integral we must integrate back with respect to $a$.
$$I(a)=\int \frac{\cos((2a)^4)-\cos(2a)}{3a}da\overset{2a=x}=\frac16 \int \frac{\cos (x^4)}{x}dx-\frac16 \int \frac{\cos x}{x}dx$$
And by using the cosine integral the second one is directly the representation of it. Also via the substitution $x^4=t$ we obtain the same integral again (like the second one, only by a factor of $\frac14$ different).
Yes it's not that nice (not even elementarry) and I wonder why was the request to take the derivative in the exercise, because we couldve used $x^3=t$ from the start then use the representation of cosine integral to arrive at the same thing.
