Dtyjlui

Finding $displaystyle int^{frac{pi}{2}}_{0}ln(sin x)cdot sin xdx$

What I try:-> Integration by parts

assuming $displaystyle I = intln(sin x)cdot sin xdx = -ln(sin x)cdot cos x+intfrac{cos^2 x}{sin x}dx$

$displaystyle I = -ln(sin x)cdot cos x+intfrac{1-sin^2 x}{sin x}dx$

$ = -ln(sin x)cos x+lnbigg(tanfrac{x}{2}bigg)-cos x$

$ displaystyle int^{frac{pi}{2}}_{0}ln(sin x)cos xdx = bigg[-ln(sin x)cos x+lnbigg(tanfrac{x}{2}bigg)-cos xbigg]bigg|^{frac{pi}{2}}_{0}=-ln(0)+ln(0)$

but answer is $ln(2/e)$

could some explain me why I have got wrong answer,thanks

also explain me How I solve it using double integral

begin{align}
I&=int^{frac{pi}{2}}_{0}ln(sin x)cdot sin x,dx
tag{1}label{1}
end{align}

begin{align}
I&=int^{frac{pi}{2}}_{0}tfrac12ln(sin^2 x)cdot sin x,dx
tag{2}label{2}
\
&=
int^{frac{pi}{2}}_{0}
tfrac12ln(1-cos^2 x)cdot sin x,dx
tag{3}label{3}
.
end{align}

Let $t=cos x$, then we have

begin{align}
I&=tfrac12int_0^1ln(1-t^2),dt
\
&=
tfrac12int_0^1ln(1-t)+ln(1+t),dt
\
&=
left.tfrac12
(
1-t-(1-t)ln(1-t)
+(t+1)ln(t+1)-1-t
)right|_0^1
=ln2-1
.
end{align}

$logsin x$ has a well-known Fourier series:
$$ logsin x=-log 2-sum_{kgeq 1}frac{cos(2k x)}{k} $$
and for any $kinmathbb{N}^+$ we have
$$ int_{0}^{pi/2}cos(2kx)sin(x),dx = -frac{1}{(2k-1)(2k+1)}, $$
hence
$$ int_{0}^{pi/2}sin(x)logsin(x),dx = -log(2)+sum_{kgeq 1}frac{1}{(2k-1)k(2k+1)} $$
where the last series equals $-1+2log 2$ by partial fraction decomposition. It follows that
$$ int_{0}^{pi/2}sin(x)logsin(x),dx = log(2)-1 $$
as wanted.

Other answers are good but I prefer to talk about yours. You found (with a typo)
begin{align}
int_{0}^{frac{pi}{2}}ln(sin x) sin x dx
&= -ln(sin x)cos x+lnbigg(tanfrac{x}{2}bigg)color{red}{+}cos xBig|_{0}^{frac{pi}{2}} \
&= 0 + lim_{xto0}bigg(ln(sin x)cos x+lntanfrac{x}{2}bigg)-1 \
&= 0 + lim_{xto0}bigg(ln(1+cos x)-(1-cos x)lnsin xbigg)-1 \
&= ln2-1
end{align}

Here is an approach following along lines similar to your own answer. There is however a small subtlety used in the first integration by parts step.

On integrating by parts, we have
$$int_0^{frac{pi}{2}} sin x ln (sin x) , dx = (1 - cos x) ln (sin x) Big{|}_0^{pi/2} - int_0^{frac{pi}{2}} (1 - cos x) cdot frac{cos x}{sin x} , dx.$$
Note the subtlety here. Having chosen $v' = sin x$ we have used $v = 1 - cos x$, that is, a non-zero constant of integration has been selected. Doing so means one has zero at the upper and lower limits of integration.

Continuing, we have
begin{align}
int_0^{frac{pi}{2}} sin x ln (sin x) , dx &= int_0^{frac{pi}{2}} frac{-cos x + cos^2 x}{sin x} , dx\
&= int_0^{frac{pi}{2}} frac{-cos x + 1 - sin^2 x}{sin x} , dx\
&= int_0^{frac{pi}{2}} left [text{cosec} , x - cot x - sin x right ] , dx\
&= left [-ln (text{cosec} ,x + cot x) - ln (sin x) + cos x right ]_0^{pi/2}\
&= left [-ln (1 + cos x) + cos x right ]_0^{pi/2}\
&= ln 2 - 1,
end{align}
as expected.