What is the Integral of sin(x)cos(x)?

Question

To be more precise, HOW do you integrate it? I know that it works out to -1/2 cos^2(x)

Alright, now how about Sin(x)Cos(c*x)dx
(thanks everyone so far...)

Answer 1

Interesting...there are actually two solutions. Let's use substitution:

Let u = cos(x). du = -sin(x)dx.

So Integral[sin(x)cos(x)]dx

= - Integral[udu]

= -u^2/2

= -[cos(x)]^2/2

Now, let u = sin(x). du = cos(x)dx.

So Integral[sin(x)cos(x)dx]

= Integral[udu]

= u^2/2 = [sin(x)]^2/2

Just to check, I did a definite integral on both results, and got the same value for both.

Pretty neat...

Answer 2

As mentioned above, a u-substitution works well. A different method is to use the double angle formula: sin2x=2(sinx)(cosx)

The integral becomes (1/2)sin2x. This integrates to -(1/4)cos2x + C.
(C is the constant of integration.)

cos2x=2cos^2(x) - 1
Thus you can rewrite the answer as -(1/4)(2cos^2(x)-1) + C = -(1/2)cos^2(x) + C'.
Here, that -1 is incorporated into C and becomes C'.

If you're wondering why all the other people are getting sin^2's, another formula for cos2x is:
cos2x=1-2sin^2(x)
Then the answer can be rewritten -(1/4)(1-2sin^2(x)) + C = (1/2)(sin^2(x)) + C'',
where the 1 is incorporated into C and becomes C''.

Answer 3

OK. Make a substitution u = sin x, du = cos x dx.
So the integral is now int[udu] = u²/2 = 1/2 sin ^2 x.+ C.
To get your answer use sin²x = 1 - cos² x and
it falls right out.
Or
You could have let u = cos x, du = -sin x dx
to get your answer.

Answer 4

Let u=sin(x)
u'=cos(x)dx
u u'=sin(x) cos(x)dx
So integrated it=(1/2) (sinx)^2+ c or you let
u=cos(x)
u'= - sin(x)dx
so -u u'=sin(x)cos(x)dx
Integrated it:= -(1/2)(cosx)^2+ c completed.

Answer 5

putting sinx=t
cosxdx=dt
integral tdt=t^2/2+C
=(1/2)sin^2x+C

Answer 6

as derivative of cosx is -sinx
therefore put
cosx=t
-sinxdx=dt
I=fsinxcosxdx=-f tdt=-t^2/2
=(-cosx^2)/2
hence the answer
thanks