Question

$\displaystyle \int \limits_{0}^{ \frac{ \pi}{2} } \tt\tan (x) \ln ( \sin (x))$
Solve The Limit;)​

Answer 1

Step-by-step explanation:

☼︎Given

$\displaystyle \int \limits_{0}^{ \frac{ \pi}{2} } \tt\tan (x) \ln ( \sin (x))$

☼︎To Find

• Solve The Limit

☼︎Solution

Let x =$\arcsin(y)(y),$so,That$\sf\sin(x)$

$\tan(x)=\dfrac y{\sqrt{1-y^2}}tan(x)=1−y2ydx = \dfrac{dy}{\sqrt{1-y^2}}$

Then the integral transforms to

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}$

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \,$

Integrate by parts, taking

u =

$\ln(y) \implies du = \dfrac{dy}yu=ln(y) = \dfrac{y}{1-y^2} \, dy$

$\implies v = -\dfrac12$

1-y^2|dv=1−y2ydy⟹v=−21ln∣1−y2∣

For 0 < y < 1, we have |1 - y²| = 1 - y², so

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \,$

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \,$

$\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n$

Recall the Taylor series for ln(1 + y),

Replacing y with -y² gives the Taylor series

$\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}$

and replacing ln(1 - y²) in the integral with its series representation gives

$\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n$

Interchanging the integral and sum (see Fubini's theorem) gives

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \,$

Compute the integral:

$</p><p>\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}$

and we recognize the famous sum (see Basel's problem),

$\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6$

So, the value of our integral is

$\footnotesize\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx =$

$\boxed{- \sf\dfrac{\pi^2}{24}} \footnotesize$

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$\huge \pink{ \blue{} \red{❀} }$тнαηк үσυ!!$\huge \pink{ ❄︎}$