# find `int (25cosx+15)/(3cosx+4sinx+5) dx`

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int (25cos x)/(3cos x+4sin x+5)dx + int (15)/(3cos x+4sin x+5)dx

The only useful method of integration right now seems to be substitution `tan (x/2) = y` and identities `sin x = (2tan(x/2))/(1 + tan^2(x/2)), cosx = (1 - tan^2(x/2))/(1 + tan^2(x/2))`

Differentiate `tan (x/2) = y:`

`(1 + tan^2(x/2))/2 dx = dy`

Make the changes suggested:

`int (25cos x+ 15)/(3cos x+4sin x+5)dx = int (25(1 - y^2)/(1 + y^2) + 15)/(3(1 - y^2)/(1 + y^2) + 4*(2y)/(1 + y^2) + 5)*(2dy)/(1 + y^2)`

Work the numerator and denominator appart:

`25(1 - y^2)/(1 + y^2) + 15 =(25 - 25y^2 + 15 + 15y^2)/(1 + y^2)`

`25(1 - y^2)/(1 + y^2) + 15 =(40 - 10y^2)/(1 + y^2)`

`3(1 - y^2)/(1 + y^2) + 4*(2y)/(1 + y^2) + 5 = (3 - 3y^2 + 8y + 5 + 5y^2)/(1 + y^2) `

`3(1 - y^2)/(1 + y^2) + 4*(2y)/(1 + y^2) + 5 = (2y^2 + 8y + 8)/(1 + y^2)`

Put together the numerator and denominator:

`((40 - 10y^2)/(1 + y^2))/((2y^2 + 8y + 8)/(1 + y^2))`

Make simplifications of common denominators:

`((40 - 10y^2)/(2y^2 + 8y + 8)) = (20 - 5y^2)/(y^2 + 4y + 4)`

`(20 - 5y^2)/(y^2 + 4y + 4) ` = 5`(4 - y^2)/(y + 2)^2`

5`(4 - y^2)/(y + 2)^2` = -5`(y^2 - 4)/(y + 2)^2`

`-5(y^2 - 4)/(y + 2)^2 = -5(y - 2)(y + 2)/(y + 2)^2`

`-5(y - 2)(y + 2)/(y + 2)^2 = -5(y - 2)/(y + 2)`

Therefore, the new integrand looks like this -5(y - 2)/(y + 2). You can notice the simplified form and much more easy to work with.

`int -5(y - 2)/(y + 2)*(2dy)/(1 + y^2) = -10 int (y - 2)/((y + 2)(1 + y^2)) dy`

Break down integrand into partial fractions:

`(y - 2)/((y + 2)(1 + y^2)) = A/(y+2) + (By+C)/(1+y^2)`

`y - 2 = A + Ay^2 + By^2 + Cy + 2By + 2C`

`y - 2 = (A+B)y^2 + (C+2B)y + A+2C`

Set equal matching parts

`(A+B)y^2 = 0 => A+B=0=>A=-B`

`(C+2B)y = y => C+2B=1`

`A+2C=-2`

Put -B in place of A in A+2C=-2 so 2C - B = -2

Create a system with equations `2C - B = -2` and `C+2B=1:`

`C = 1-2B => 2(1 - 2B) - B = -2`

`2 - 4B - B = -2 => -5B = -4 => B = 4/5 => A = -4/5 => C = 1 - 8/5 => C = -3/5`

`(y - 2)/((y + 2)(1 + y^2)) = -4/(5(y+2)) + (4y-3)/(5(y^2+1))`

Break down integral in two:

`-10 int (y - 2)/((y + 2)(1 + y^2)) dy = 8int 1/(y+2)dy - 2 int (4y-3)/((y^2+1))` `-10 int (y - 2)/((y + 2)(1 + y^2)) dy = 8int 1/(y+2)dy - 4 int (2y)/(y^2+1)dy +2int 3/(1+y^2)dy`

`-10 int (y - 2)/((y + 2)(1 + y^2)) dy = 8 ln |y+2| - 4 ln(y^2+1) + 6 arctan y + c`

Do not forget that `y = tan (x/2)`

`int (25cos x+ 15)/(3cos x+4sin x+5)dx = 8ln |tan (x+2) + 2| - 4 ln (tan^2(x/2) + 1) + 6 arctan (tan (x/2)) + c` **Your answer : `int (25cos x+ 15)/(3cos x+4sin x+5)dx = 8ln |tan (x+2) + 2| - 4 ln (tan^2(x/2) + 1) + 6 arctan (tan (x/2)) + c` **

This can be solved using the following method.

We will define the numerator in terms of denominator and its derivative.

`25cosx + 15 = A (3cosx +4sinx+5)+B(-3sinx+4cosx)+C`

Comparing coefficients

`cosrarr25=3A+4B ---(1)`

`sinrarr0=4A-3B ---(2)`

`const.rarr15=5A+C ---(3)`

Solving (1),(2) and (3) will give you;

A=3

B=4

C=0

`int(25cosx + 15)/(3cosx +4sinx+5) dx `

`= int (3(3cosx+4sinx+5)+4(-3sinx+4cosx))/(3cos+4sinx+5)dx`

`= 3 int dx +4 int (-3sinx+4cosx)/(3cosx+4sinx+5)dx`

`=3x+4 ln(3cosx+4sinx+5)+K ` where K is a constant.

*So the answer is;*

`int(25cosx + 15)/(3cosx +4sinx+5) dx= 3x+4 ln(3cosx+4sinx+5)+K`

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