0-pi-x-ln-sin-x-dx-

Question Number 82174 by jagoll last updated on 18/Feb/20

\underset{0}{\int^{π}} x l n (\sin x) d x = ?

Commented by mathmax by abdo last updated on 19/Feb/20

l e t I = \int_{0}^{π} x l n (s i n x) d x \Rightarrow I = \int_{0}^{\frac{π}{2}} x l n (s i n x) d x + \int_{\frac{π}{2}}^{π} x l n (s i n x) d x

b u t \int_{\frac{π}{2}}^{π} x l n (s i n x) d x =_{x = \frac{π}{2} + α} \int_{0}^{\frac{π}{2}} (\frac{π}{2} + α) l n (c o s α) d α

= \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α + \int_{0}^{\frac{π}{2}} α l n (c o s α) d α \Rightarrow

I = \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α + \int_{0}^{\frac{π}{2}} x (l n (s i n x) + l n (c o s x) d x

= \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α + \int_{0}^{\frac{π}{2}} x l n (\frac{1}{2} s i n (2 x)) d x

= \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α - l n (2) \int_{0}^{\frac{π}{2}} x d x + \int_{0}^{\frac{π}{2}} x l n (2 x) {d x}_{\to 2 x = t}

= \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α - l n (2) {[\frac{x^{2}}{2}]}_{0}^{\frac{π}{2}} + \frac{1}{2} \int_{0}^{π} t l n (t) \frac{d t}{2}

= \frac{π}{2} \int_{0}^{\frac{π}{2}} l n (c o s α) d α - l n (2) {\frac{π^{2}}{8}} + \frac{1}{4} I \Rightarrow

\frac{3}{4} I = \frac{π}{2} (- \frac{π}{2} l n (2)) - \frac{π^{2}}{8} l n (2) = - \frac{π^{2}}{4} l n (2) - \frac{π^{2}}{8} l n (2)

= - \frac{3 π^{2}}{8} l n (2) \Rightarrow I = \frac{4}{3} (- \frac{3 π^{2}}{8}) l n (2) = - \frac{π^{2}}{2} l n (2)

Answered by Kunal12588 last updated on 19/Feb/20

I = \int_{0}^{π} x l o g (s i n x) d x

\Rightarrow I = \int_{0}^{π} (x - π) l o g (s i n x) d x

\Rightarrow I = \frac{π}{2} \int_{0}^{π} l o g (s i n x) d x

\Rightarrow I = π \int_{0}^{π / 2} l o g (s i n x) d x

[★ \int_{0}^{π / 2} l o g (s i n x) d x = \int_{0}^{π / 2} l o g (c o s x) d x = - \frac{π}{2} l o g (2)]

\Rightarrow I = π (\frac{- π}{2} l o g (2)) = \frac{- π^{2}}{2} l o g (2)

∴ \int_{0}^{π} x l o g (s i n x) d x = - \frac{π^{2}}{2} l o g (2)

[★ h e r e “ l o g a^{″} m e a n s “ {l o g}_{e} a^{″}]

Commented by Kunal12588 last updated on 19/Feb/20

★ T . P \int_{0}^{π / 2} l o g (s i n x) d x = \frac{- π}{2} l o g (2)

I = \int_{0}^{π / 2} l o g (s i n x) d x

\Rightarrow I = \int_{0}^{π / 2} l o g (c o s x) d x

\Rightarrow I = \frac{1}{2} \int_{0}^{π / 2} l o g (s i n x c o s x) d x

\Rightarrow I = \frac{1}{2} \int_{0}^{π / 2} l o g (s i n 2 x) d x - \frac{1}{2} \int_{0}^{π / 2} l o g 2 d x

\Rightarrow I = \frac{1}{2} \int_{0}^{π / 2} l o g (s i n 2 x) d x - \frac{π}{4} l o g 2

l e t 2 x = t \Rightarrow d x = \frac{1}{2} d t

I = \frac{1}{4} \int_{0}^{π} l o g (s i n t) d t - \frac{π}{4} l o g 2

\Rightarrow I = \frac{1}{2} \int_{0}^{π / 2} l o g (c o s x) d x - \frac{π}{4} l o g 2

\Rightarrow I = \frac{1}{2} \int_{0}^{π / 2} l o g (s i n x) d x - \frac{π}{4} l o g 2

\Rightarrow I = \frac{1}{2} I - \frac{π}{4} l o g 2

\Rightarrow \frac{1}{2} I = - \frac{π}{4} l o g 2

\Rightarrow I = - \frac{π}{2} l o g 2

\Rightarrow \int_{0}^{π / 2} l o g (s i n x) d x = \frac{- π}{2} l o g (2)