I'm at a loss. How do I integrate?

Question

Integrate (7+10cot(x))/(10-7cot(x)) from pi/4 to pi/2

Answer 1

(7+10cot(x))/(10-7cot(x))

 

(7 + 10cosx/sinx) / (10 - 7cosx/sinx)

 

(7sinx +10cosx) / (10sinx - 7cosx)     the sinx in both denominators drops out

 

let u = 10sinx -7cosx

 

du = 10cosx + 7sinx   dx

 

∫ du/u = ln (10sinx - 7cosx) evaluated from pi/4 to pi/2

 

ln(10 - 0) - ln(3√2/2) = 1.55

Answer 2

One approach is to rewrite the integrand as

 

(7 sin(x) + 10 cos(x) )/ (10 sin(x) - 7 cos(x) )

 

Standard trig analysis will show that 7 sin(x) + 10 cos(x) = 12.2 sin( x + 55)   and

                10 sin(x) - 7 cos(x)  = -12.2 cos(x + 55)

  {  the 12.2 is an approximation for  sqrt(149)  and 55 deg is a very good approximation for  arcsin(10/sqrt(149)  }

 

The integrand is thus  - sin(x + 55 / cos(55) =  - tan(x + 55)

 

Standard tables of integrals    show that the integral of tan = ln (|sec(x) |) + C

 

The integral is  thus   - ln( |sec( 90 + 55) |) + ln ( |sec(45 + 55) | ) 

 

This works out to 1.55