Question

How do I find the integral `intcos(x)ln(sin(x))dx` ?

Answer 1

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

Explanation

`=intcos(x)ln(sin(x))dx`

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

`=intln(t)*1dt`

Using Integration by Parts,

`int(I)(II)dx=(I)int(II)dx-int((I)'int(II)dx)dx`

where `(I)` and `(II)` are functions of `x`, and `(I)` represents which will be differentiated and `(II)` will be integrated subsequently in the above formula

Similarly following for the problem,

`=ln(t)int1*dt-int((lnt)'intdt)dt`

`=tln(t)-int1/t*tdt`

`=tln(t)-t+c`, where `c` is a constant

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

`=sin(x)(lnsin(x)-1)+c`, where `c` is a constant