Question

How do I find the integral `intcos(x)/(sin^2(x)+sin(x))dx` ?

Answer 1

`=ln(sinx/(sin(x)+1))+c`, where `c` is a constant

Full Answer

`=intcos(x)/(sin^2(x)+sin(x))dx`

`=intcos(x)/(sin(x)(sin(x)+1))dx`

let's `sin(x)=t`, then, `cos(x)dx=dt`

`=int1/(t(t+1))dt`

Using Partial Fractions,

`1/(t(t+1))=A/t+B/(t+1)` .............`(i)`

multiplying both sides with `t(t+1)`,

`1=A(t+1)+Bt`

`1=A+(A+B)t`

comparing constant and coefficient on both sides, we get

`A=1`,

`A+B=0`, which implies `B=-1`

plugging the values of `A` and `B` in `(i)`,

`1/(t(t+1))=1/t-1/(t+1)`

integrating both sides with respect to `t`,

`int1/(t(t+1))dt=int1/tdt-int1/(t+1)dt`

`=lnt-ln(t+1)+c`, where `c` is a constant

`=ln(t/(t+1))+c`, where `c` is a constant

substituting `t`, yields

`=ln(sinx/(sin(x)+1))+c`, where `c` is a constant