I'm sorry, I'm not sure what you're asking for. Can you please clarify?

[nameVoid]

Homework Statement

I(secx*cos2x/(sinx+secx),x)
I(cos2x/(sinxcosx+1),x)
I(cos2x/(sin2x/2+1),x)
u=sin2x du/2=cos2xdx
1/2 I( 1/ (u/2+1) , x)
I( 1/ (u+2) , x)
ln|u+2|+C=ln|sin2x+2|+C

also

I((x^2-bx)sin2x,x)
I(x^2sin2x,x)-bI(xsin2x,x)
I(x^2sin2x,x):
u=x^2 du= 2xdx
dv=sin2x v=-(1/2)cos2x
-x^2cos2x/2+I(xcos2x,x)
u=x du=dx
dv=cos2x v=sin2x/2
-x^2cos2x/2+xsin2x/2-I(sin2x,x)/2
-x^2cos2x/2+xsin2x/2+cos2x/4

-bI(xsin2x,x):
u=x du=dx
dv=sin2x v=-cos2x/2
-b[-xcos2x/2+I(cos2x,x)/2]
bxcos2x/2-bsin2x/4

thus:
I((x^2-bx)sin2x,x)=-x^2cos2x/2+xsin2x/2+cos2x/4+bxcos2x/2-bsin2x/4+C

Homework Equations

The Attempt at a Solution

 

[mighty2000]

In order to solve this integral, we can use the substitution method. Let u = sin2x, then du = 2cos2x dx. We can rewrite the integral as:

I(secx*cos2x/(sinx+secx),x) = I(1/(u+1),x)

Next, we can use the formula for integrating 1/(x+a), which is ln|x+a|+C. Therefore, the final solution is:

I(secx*cos2x/(sinx+secx),x) = ln|sin2x+2|+C

For the second integral, we can use integration by parts. Let u = x^2 and dv = sin2x, then du = 2x dx and v = -(1/2)cos2x. Therefore, the integral becomes:

I((x^2-bx)sin2x,x) = -x^2cos2x/2 + xsin2x/2 - I(xcos2x,x)/2

Next, let u = x and dv = cos2x, then du = dx and v = sin2x/2. Thus, the integral becomes:

I((x^2-bx)sin2x,x) = -x^2cos2x/2 + xsin2x/2 + cos2x/4 + bxcos2x/2 - bsin2x/4 + C

And that is our final solution for the second integral.