How to prove correctness of a formula (differential calculus, integral)? How do I prove the correctness of the following formula relating to the fundamental theorem of calculus?
$$\int \! x\cos{3x} \, \mathrm{d}x = \frac{\cos{3x}}{9}+\frac{x\sin{3x}}{3}+C$$
 A: Differentiation is easy!  Once you have calculated an integral, differentiate your "answer" and see whether you get the right thing.
Of course you have to remember the $+{}C$, since if you differentiate, you will get the same thing as if you had remembered it.
You can even use the idea as a "technique of integration."
Here is a simple example.
We want $\int e^{3x}dx$.  Guess that the answer is $e^{3x}+C$. 
Now differentiate your answer.  If you remember to use the Chain Rule, you will get $3e^{3x}$.  So the answer $e^{3x}+C$ was wrong.  But that's easy to fix.  Divide your "answer" $e^{3x}$ by $3$ to get rid of that nasty $3$ that came from the Chain Rule.  You get a new improved answer $\frac{e^{3x}}{3}+C$.  If you still have doubts, differentiate that. You will get $e^{3x}$, the thing you were trying to integrate.
Even on a test, if there is time, you can check all your (indefinite) integral answers by differentiating.  You probably have roughly $100$ percent mastery of differentiation, and can do it quickly, so a check is always possible, and quick.
A: Integration by parts would be of great help. Let $u = x$, and let $dv = \cos{3x} \:\rm{dx}$. Then you have $du = dx$ and $v = \frac{\sin{3x}}{3}$. Using the formula we have $$I = \int x \cos{3x} \ \text{dx} = uv - \int v \ \text{du} =  \frac{x \cdot\sin{3x}}{3} - \frac{1}{3}\int\sin{3x} \ \text{dx}$$
To see whether your answer is correct differentiate the Right Hand side. So you have $$ \frac{1}{9} \cdot -3\sin{3x} + \frac{1}{3} \sin{3x} + x \cos{3x} = x \cos{3x}$$
