Question

How do you find the indefinite integral of `int (cot x)dx/(1+cos x)dx`?

Answer 1

`I=ln|cscx-cotx|-1/4ln|(cosx-1)/(cosx+1)|+1/(2(1+cosx))+C`

Explanation:

Here,

`I=intcotx/(1+cosx)dx`

`=intcosx/(sinx(1+cosx))dx`

`=int((1+cosx)-1)/(sinx(1+cosx))dx`

`=int1/sinxdx-int1/(sinx(1+cosx))dx`

`=intcscxdx-intsinx/(sin^2x(1+cosx))dx`

`=ln|cscx-cotx|-intsinx/((1-cos^2x)(1+cosx))dx`

`I=ln|cscx-cotx|-I_1,...to(D)`

`where,` `I_1=intsinx/((1-cosx)(1+cosx)^2)dx`

Let, `cosx=u=>sinxdx=-du`

So,

`I_1=int(-1)/((1-u)(1+u)^2)du`

Now,

`(-1)/((1-u)(1+u)^2)=A/(1-u)+B/(1+u)+C/(1+u)^2`

`:. -1=A(1+u)^2+B((1-u)(1+u))+C(1-u)`

`u=1=>-1=A(1+1)^2=>color(red)(A=-1/4`

`u=-1=>-1=C(1+1)=>color(red)(C=-1/2`

`u=0=>-1=A(1)+B(1)+C(1)=>color(red)(A+B+C=-1`

`=>(-1/4)+B+(-1/2)=-1`

`=>color(red)(B=1/4+1/2-1=-1/4`

Thus,

`I_1=int(-1/4)/(1-u)du+int(-1/4)/(1+u)du+int(-1/2)/(1+u)^2du`

`I_1=1/4int1/(u-1)du-1/4int1/(1+u)du-1/2int(1+u)^(-2) du`

`=1/4ln|u-1|-1/4ln|u+1|-1/2((1+u)^-1/(-1))+c`

`=1/4ln|(u-1)/(u+1)|+1/2*1/(1+u)+c,where,u=cosx`

`:.I_1=1/4ln|(cosx-1)/(cosx+1)|+1/(2(1+cosx))+c`

Hence, from `(D)`

`I=ln|cscx-cotx|-1/4ln|(cosx-1)/(cosx+1)|+1/(2(1+cosx))+C`