Intergration by parts for sin(x)cos(x)

[uzman1243]

I know its easier to use the substitution method, by I'm trying to see how it'll work for integration by parts. I follow the LIATE method for integration by parts.

Now if I take u=cos(x) and dv = sin(x), the answer changes.

Can you please explain this to me? Which is the 'right' answer?

 

[gopher_p]

Rather than answer your question directly, let me provide another example of the same phenomenon that might make it easier to identify why both answers are correct.

If you compute $\int (x+1)\ dx$ using the sum rule for antiderivatives and the "reverse power rule", you get $$\int x+1\ dx=\frac{1}{2}x^2+x+C.$$ If instead you use a $u$-sub, with $u=x+1$, you get $$\int (x+1)\ dx=\int u\ du=\frac{1}{2}u^2+C=\frac{1}{2}(x+1)^2+C.$$

Now the exact same thing is happening in your example as is happening here; you have two different-looking answers that are both correct. How can that be?

 

[D H]

And closer to home, let's do $\int \sin(x) \cos(x) dx$ a third way, using the trig substitution $\sin(2x)=2\sin(x) \cos(x)$. Thus $\int \sin(x) \cos(x) dx = \int \frac {\sin(2x)}2 dx = -\frac {\cos(2x)} 4 + C$.

To quote gopher_p,

How can that be?

 

[athosanian]

dear friend , all these answers are different only by a constant, so they are the same answers.