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

1. May 30, 2014

### 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?

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2. May 30, 2014

### 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?

3. May 30, 2014

### D H

Staff Emeritus
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,

4. May 30, 2014

### athosanian

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