Question

Integrate :
sin(x)cos(x)dx

Answer 1

If we use u-substituition then,

let sin(x) u
du =cos(x)dx

that gives us:
u^2/2 + C
sin^x/2 + C

Approach 2:

let cos(x) = u'
u'=cos(x)
du' = -sin(x)dx
that gives us:
u'^2/2 + C
-cos^2(x)/2 + C

Approach 3

Using sin2(x) = 2sin(x)cos(x)

will give:

-cos2(x)/4+C

Answer 2

Now my question is,
how can we get 3 different answers for the same question using 3 seemingly right ways?

Answer 3

In the last approach ,
I've skipped the step in which we multiply both sides by 1/2.

Answer 4

\[\large\rm -\frac14\cos(2x)+c\]Applying Cosine Double Angle Identity gives us,\[\large\rm -\frac14(1-2\sin^2x)+c\]Distributing,\[\large\rm \frac12\sin^2x-\frac14+c\]Absorb the -1/4 into the constant to create a new constant,\[\large\rm \frac12\sin^2x+C\]Understand how the third answer and first are actually the same? :)

I'll bet the second answer lines up in this way through some trig identities.

Answer 5

Interesting.

It seems like all the three answers are connected by means by a variety of trigonometric function and identities.

Answer 6

Ya it's interesting that you can get different answer based on the method you used. Trig is so crazy :P

Answer 7

Yes!
Thats what I really like about trig:P

Now ,do you mind explaining how does the second answer line up with the other two ?

Answer 8

Like,

is
ans1=ans2=ans3?

Answer 9

\[\large\rm -\frac12\cos^2x+c\]Apply your Pythagorean Identity,\[\large\rm -\frac12(1-\sin^2x)+c\]Then same idea from there,
distribute,\[\large\rm \frac12\sin^2x-\frac12+c\]absorb into constant,\[\large\rm \frac12\sin^2x+C\]

Answer 10

Yes, they are the same, because of the constant.
The arbitrary nature of the constant allows us to line things up exactly,
doing this fancy absorption business.

Answer 11

Seems like :
1/2 + C is still a constant...

Answer 12

Locking 1/2 +C is still C?

Answer 13

Hmm I'm not sure what you mean :)
The absorbing thing is a little confusing?

Think of c as ANY NUMBER.
If you add 1/2 to that value, it doesn't change anything, right?
c could still take on any value.

So we'll call c+1/2 a new constant which can be ANY NUMBER.
Maybe call it m if lowercase and capital c was not clear.

Answer 14

I wasn't saying 1/2 + C = C.
I was saying 1/2 + c = C.

I was using lowercase c for the old constant.

Answer 15

I know that.
I was taking about the nature of the constant :P