bemath-0-2pi-ln-1-sin-x-dx-

Question Number 107314 by bemath last updated on 10/Aug/20

⩞ b e m a t h ⩞

\underset{0}{\int^{2 π}} \ln (1 + \sin x) d x ?

Answered by mnjuly1970 last updated on 10/Aug/20

Ω = \int_{0}^{2 π} l n (1 + s i n (x)) d x = \int_{- π}^{π} l n (1 - s i n (x)) d x

2 Ω = 2 \int_{0}^{π} l n (c o s (x)) d x \Rightarrow Ω = - π l n (2) \dots

Commented by bemath last updated on 10/Aug/20

t h a n k y o u

Answered by mathmax by abdo last updated on 10/Aug/20

\int_{0}^{2 π} \ln (1 + sinx) dx = \int_{0}^{π} \ln (1 + sinx) dx + \int_{π}^{2 π} \ln (1 + sinx) dx (\to x = π + t)

= \int_{0}^{π} \ln (1 + sinx) dx + \int_{0}^{π} \ln (1 - sint) dt = \int_{0}^{π} \ln {(1 + sinx) (1 - sinx)} dd

= \int_{0}^{π} \ln (1 - \sin^{2} x) dx = \int_{0}^{π} \ln (\cos^{2} x) dx = 2 \int_{0}^{π} \ln ∣ cosx ∣ dx

= 2 \int_{0}^{\frac{π}{2}} \ln (cosx) dx + 2 \int_{\frac{π}{2}}^{π} \ln (- cosx) dx but

\int_{0}^{\frac{π}{2}} \ln (cosx) dx = - \frac{π}{2} \ln (2) and

\int_{\frac{π}{2}}^{π} \ln (- cosx) dx = \int_{\frac{π}{2}}^{π} \ln (\cos (π - x)) dx =_{π - x = t} \int_{\frac{π}{2}}^{0} \ln (cost) (- dt)

= \int_{0}^{\frac{π}{2}} \ln (cost) dt = - \frac{π}{2} \ln (2) \Rightarrow I = - π \ln (2) - π \ln (2) \Rightarrow

I = - 2 π \ln (2)