Question

Integral of (12sinx-2cosx+3)/(sinx+cosx) ?

Answer 1

`I=5x-7ln(cos(x)+sin(x))+6/sqrt(2)tanh^-1((tan(x/2)-1)/sqrt(2))+C`

Explanation:

We want to solve

`I=int(12sin(x)-2cos(x)+3)/(sin(x)+cos(x))`

Let's split this into two separate integral

`I=int(12sin(x)-2cos(x))/(sin(x)+cos(x))dx+int3/(sin(x)+cos(x))dx`

First integral

Let

`I_1=int(12sin(x)-2cos(x))/(sin(x)+cos(x))dx`

Consider the much easier integral

`I_M=Aint(sin(x)+cos(x))/(sin(x)+cos(x))dx+Bint(cos(x)-sin(x))/(sin(x)+cos(x))dx`

Now determinate the constants `color(red)(A` and `color(red)(B`, such that `color(blue)(I_1=I_M`

`A-B=12` and `A+B=-2`

Thus `color(red)(A=5` and `color(red)(B=-7`, so

`I_1=5int(sin(x)+cos(x))/(sin(x)+cos(x))dx-7int(cos(x)-sin(x))/(sin(x)+cos(x))dx`

`color(white)(I_1)=5x-7ln(cos(x)+sin(x))+C_1`

Second integral

Let

`I_2=int3/(sin(x)+cos(x))dx`

Substitute `color(blue)(u=tan(x/2) larr color(red)("Weierstrass substitution"`

`I_2=6int1/((2u)/(1+u^2)+(1-u^2)/(1+u^2))*1/(u^2+1)du`

`color(white)(I_2)=6int1/(-u^2+2u+1)du`

`color(white)(I_2)=6int1/(2-(u-1)^2)du`

Let `color(blue)(s=u-1=>ds=du`

`I_2=6int1/(2-s^2)ds`

`color(white)(I_2)=3int1/(1-(s/sqrt(2))^2)ds`

`color(white)(I_2)=6/sqrt(2)tanh^-1(s/sqrt(2))+C_1`

Substitute back `color(blue)(s=u-1` and `color(blue)(u=tan(x/2)`

`I_2=6/sqrt(2)tanh^-1((tan(x/2)-1)/sqrt(2))+C_1`

Combining these

`I=5x-7ln(cos(x)+sin(x))+6/sqrt(2)tanh^-1((tan(x/2)-1)/sqrt(2))+C`

Hopefully not too many typos :-)