Integration frm FITZPATRIK 24g q9 !! >.<" (1 Viewer)

[jesshika]

How would i integrate


sin x/2 dx


The book doesn't give a let u = something
so im abit confused >.<"

 

[CM_Tutor]

The book doesn't give a substitution as it isn't a substitution problem. You need to rewrite it as a cos2@, as follows. (Note that this is the standard approach for integrating sin² and cos² functions)

Note that cos2@ = 1 - 2sin²@

We want a sin²(x / 2), so put @ = x / 2, and rearrange to make sin²@ the subject, ie:
sin²(x / 2) = (1 - cos x) / 2

So, int sin²(x / 2) dx = int (1 - cos x) / 2 dx = (1 / 2) int 1 - cos x dx
= (1 / 2) [x - sin x] + C, for some constant C
= (x - sin x) / 2 + C

 

[jesshika]

ohhhhh ...
>., i c i c ...

thankyou cm tutor ..


what about

Int (1 + cos2x)sinx dx

do i subsitute 1-2sinx in the above ???
and integrate from there ?

 

[CM_Tutor]

Originally posted by jesshika
what about

Int (1 + cos2x)sinx dx

do i subsitute 1-2sinx in the above ???
and integrate from there ?

You could substitute cos 2x = 2cos²x - 1, and then do the problem using the substitution u = cos x.

Alternately, you could expand to get a cos 2x * sin x term, and then use the products to sums formula
cos a * sin b = [sin(a + b) - sin(a - b)] / 2, with a = 2x and b = x.
Thus, cos 2x * sin x = [sin(2x + x) - sin(2x - x)] / 2 = (sin 3x - sin x) / 2
So, int (1 + cos 2x)sin x dx = int sinx + (sin 3x - sin x) / 2 dx = (1 / 2) int sin 3x + sin x dx

Since both methods work, and only differ by a constant, you could then find constants A, B and C such that
cos³x = Acos 3x + Bcos x + C - this could be a good exam question.

Note: There are other ways to complete this problem.

 

[jesshika]

thanx cm tutor ~~