$\int_{0}^{π} log ({sin}^{2} x) d x = 0$ $int_0^pilog(sin^2 x) dx=0$ ?

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$\int_{0}^{π} log ({sin}^{2} x) d x = 0$ $int_0^pilog(sin^2 x) dx=0$ ?

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Answer 1 · 2018-04-12T09:32:21

$I = - ln (4) π$ $I=-ln(4)pi$

Explanation:

We want to solve

$I = \int_{0}^{π} ln ({sin}^{2} (x)) d x$ $I=int_0^piln(sin^2(x))dx$

By the properties of logarithms

$I = 2 \int_{0}^{π} ln (sin (x)) d x$ $I=2int_0^piln(sin(x))dx$

Use the trig identity

$sin (x) = 2 sin (\frac{x}{2}) cos (\frac{x}{2})$ $color(blue)(sin(x)=2sin(x/2)cos(x/2)$

Thus

$I = 2 \int_{0}^{π} ln (2 sin (\frac{x}{2}) cos (\frac{x}{2})) d x$ $I=2int_0^piln(2sin(x/2)cos(x/2))dx$

$= 2 \int_{0}^{π} ln (2) d x + 2 \int_{0}^{π} ln (sin (\frac{x}{2})) d x + 2 \int_{0}^{π} ln (cos (\frac{x}{2})) d x$ $=2int_0^piln(2)dx+2int_0^piln(sin(x/2))dx+2int_0^piln(cos(x/2))dx$

$= ln (4) π + 2 \int_{0}^{π} ln (sin (\frac{x}{2})) d x + 2 \int_{0}^{π} ln (cos (\frac{x}{2})) d x$ $=ln(4)pi+2int_0^piln(sin(x/2))dx+2int_0^piln(cos(x/2))dx$

Make a substitution $u = \frac{x}{2} \Rightarrow d u = \frac{1}{2} d x$ $color(red)(u=x/2=>du=1/2dx$
$x = 0 \Rightarrow u = 0$ $color(red)(x=0=>u=0$ and $x = π \Rightarrow u = \frac{π}{2}$ $color(red)(x=pi=>u=pi/2$

$I = ln (4) π + 4 \int_{0}^{\frac{π}{2}} ln (sin (u)) d u + 4 \int_{0}^{\frac{π}{2}} ln (cos (u)) d u$ $I=ln(4)pi+4int_0^(pi/2) ln(sin(u))du+4int_0^(pi/2) ln(cos(u))du$

Make a substitution $u = - \frac{π}{2} - s \Rightarrow d u = - d s$ $color(red)(u=-pi/2-s=>du=-ds$
$u = 0 \Rightarrow s = - \frac{π}{2}$ $color(red)(u=0=>s=-pi/2$ and $u = \frac{π}{2} \Rightarrow s = - π$ $color(red)(u=pi/2=>s=-pi$

$I = ln (4) π + 4 \int_{0}^{\frac{π}{2}} ln (sin (u)) d u - 4 \int_{- \frac{π}{2}}^{- π} ln (- sin (s)) d s$ $I=ln(4)pi+4int_0^(pi/2) ln(sin(u))du-4int_(-pi/2)^(-pi) ln(-sin(s))ds$

$I = ln (4) π + 4 \int_{0}^{\frac{π}{2}} ln (sin (u)) d u - 4 \int_{- \frac{π}{2}}^{- π} ln (sin (- s)) d s$ $color(white)(I)=ln(4)pi+4int_0^(pi/2) ln(sin(u))du-4int_(-pi/2)^(-pi) ln(sin(-s))ds$

Make a substitution $w = - s \Rightarrow d w = - d s$ $color(red)(w=-s=>dw=-ds$
$s = - \frac{π}{2} \Rightarrow w = \frac{π}{2}$ $color(red)(s=-pi/2=>w=pi/2$ and $s = - π \Rightarrow w = π$ $color(red)(s=-pi=>w=pi$

$I = ln (4) π + 4 \int_{0}^{\frac{π}{2}} ln (sin (u)) d u + 4 \int_{\frac{π}{2}}^{π} ln (sin (w)) d w$ $I=ln(4)pi+4int_0^(pi/2) ln(sin(u))du+4int_(pi/2)^(pi) ln(sin(w))dw$

$I = ln (4) π + 2 I$ $I=ln(4)pi+2I$

$I = - ln (4) π$ $I=-ln(4)pi$

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$\int_{0}^{π} log ({sin}^{2} x) d x = 0$ $int_0^pilog(sin^2 x) dx=0$ ?

1 Answer

Explanation:

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∫π0log(sin2x)dx=0int_0^pilog(sin^2 x) dx=0?

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$\int_{0}^{π} log ({sin}^{2} x) d x = 0$ $int_0^pilog(sin^2 x) dx=0$ ?