With is the integral of (9(sinx+cosx))/sin(2x)
9(sinx + cosx)/sin2x
= 9( sinx/(2sinxcosx) + cosx/(2sinxcosx)
= (9/2)(1/cosx + 1/sinx)
= (9/2) (secx + cscs)
can you take it from there?
(the integrals of secx and cscx should be part of your repertoire of basic integrals)
To find the integral of (9(sinx+cosx))/sin(2x), we can use the method of substitution.
Let's start by simplifying the expression a bit. We notice that the numerator of the integral, 9(sinx + cosx), can be written as 9sqrt(2)sin(x + pi/4). So, we can rewrite the integral as:
∫ (9sqrt(2)sin(x + pi/4)) / sin(2x) dx
Now, let's make a substitution:
Let u = x + pi/4, which implies du = dx.
We can rewrite the integral in terms of u:
∫ (9sqrt(2)sin(u)) / sin(2u - pi/2) du
Now, let's use the trigonometric identity sin(2u - pi/2) = cos(2u), and rewrite the integral once again:
∫ (9sqrt(2)sin(u)) / cos(2u) du
This expression can be simplified further using the identity sin(u)/cos(v) = tan(u-v), giving us:
∫ (9sqrt(2)tan(u-2u)) du
= ∫ (9sqrt(2)tan(-u)) du
= -9sqrt(2) ∫ tan(u) du
Finally, we can integrate -9sqrt(2)tan(u) with respect to u. The integral of tan(u) is ln(|sec(u)|), so we get:
-9sqrt(2) ln(|sec(u)|) + C
Substituting back u = x + pi/4, we have:
-9sqrt(2) ln(|sec(x + pi/4)|) + C
So, the integral of (9(sinx+cosx))/sin(2x) is -9sqrt(2) ln(|sec(x + pi/4)|) + C.