Respuesta :
Let [tex]x = \arcsin(y)[/tex], so that
[tex]\sin(x) = y[/tex]
[tex]\tan(x)=\dfrac y{\sqrt{1-y^2}}[/tex]
[tex]dx = \dfrac{dy}{\sqrt{1-y^2}}[/tex]
Then the integral transforms to
[tex]\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}[/tex]
[tex]\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy[/tex]
Integrate by parts, taking
[tex]u = \ln(y) \implies du = \dfrac{dy}y[/tex]
[tex]dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|[/tex]
For 0 < y < 1, we have |1 - y²| = 1 - y², so
[tex]\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy[/tex]
It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have
[tex]\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy[/tex]
Recall the Taylor series for ln(1 + y),
[tex]\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n[/tex]
Replacing y with -y² gives the Taylor series
[tex]\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}[/tex]
and replacing ln(1 - y²) in the integral with its series representation gives
[tex]\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy[/tex]
Interchanging the integral and sum (see Fubini's theorem) gives
[tex]\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy[/tex]
Compute the integral:
[tex]\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}[/tex]
and we recognize the famous sum (see Basel's problem),
[tex]\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6[/tex]
So, the value of our integral is
[tex]\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \boxed{-\frac{\pi^2}{24}}[/tex]
We have,
[tex]\displaystyle \int \limits_{0}^{ \frac{ \pi}{2} } \tt\tan (x) \ln ( \sin (x))\\ \\\displaystyle \sf{ \implies \: I = \int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin \left( \dfrac{\pi}{2} - x \right) \right) \: dx } \\ \\\displaystyle \sf{ \implies \: I = \int^{ \frac{\pi}{2} }_{0} \: \ln \left( cos(x) \right) \: dx } \\ \\\displaystyle \sf{ \implies \:2I =\int^{ \frac{\pi}{2} }_{0} \:\ln \left( sin(x) \right) \: dx + \int^{ \frac{\pi}{2} }_{0} \: \ln \left( cos(x) \right) \: dx } \\ \\\displaystyle \sf{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin(x) \right) +\ln\left( cos(x) \right) \: dx }[/tex]
[tex]\displaystyle \sf{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin(x) \: cos(x) \right) \: dx } \\ \\\displaystyle\sf{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \: \ln \left( \dfrac{sin(2x)}{2}\right) \: dx } \\ \\\displaystyle \sf{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \:\ln \left( sin(2x) \right)\:dx-\ln(2) \int^{ \frac{\pi}{2} }_{0} \: dx}[/tex]
Put 2x = t, so,
[tex]\displaystyle\sf{ \implies \: 2I = \dfrac{1}{2} \int^{ \pi}_{0} \: \ln \left( sin(t) \right) \:dt - \ln(2) \int^{ \frac{\pi}{2} }_{0} \: dx} \\ \\\displaystyle\tt{ \implies \: 2I = \dfrac{1}{2} \cdot2 \int^{ \frac{\pi}{2}}_{0} \: \ln \left( sin(t) \right) \: dt - \ln(2) \int^{ \frac{\pi}{2} }_{0} \: dx}[/tex]
[tex]\displaystyle \tt{ \implies \: 2I =\int^{ \frac{\pi}{2}}_{0} \: \ln \left( sin(t) \right) \: dt - \ln(2)\int^{ \frac{\pi}{2} }_{0} \: dx} \\ \\ \displaystyle \tt{ \implies \: 2I = I - \ln(2) \left[x \right] ^{ \frac{\pi}{2} }_{0} } \\ \\ \displaystyle \tt{ \implies \: I = -\ln(2)\left[ \dfrac{\pi}{2} - 0 \right] }[/tex]
[tex]\displaystyle \sf{ \implies \: I = - \dfrac{\pi}{2} \ln(2) }[/tex]
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☞︎︎︎Apologies,if incorrect.
