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I am trying to solve the following problem:

Let $$A=\pi^2\int_{0}^{1}\frac{\sin(\pi x)}{1 + \sin(\pi x)}dx \quad \text{and}\quad B=\int_{0}^{\pi}\frac{x\sin( x)}{1 + \sin( x)}dx.$$ Find the value of $\dfrac{A}{B}$.

I thought on using integration by parts.

It is clear that $$A=\pi^2\int_{0}^{1}\frac{\sin(\pi x)}{1 + \sin(\pi x)}dx = \pi\int_{0}^{\pi}\frac{\sin( x)}{1 + \sin( x)}dx.$$ If we consider $f(x) = \dfrac{\sin( x)}{1 + \sin( x)}$, we have that $A = \pi\int_{0}^{\pi}f(x)dx$ and $B = \int_{0}^{\pi}xf(x)dx$. Applying integration by parts, we obtain $$B = \int_{0}^{\pi}xf(x)dx = x^2f(x)\Big|_0^\pi - \int_{0}^{\pi} x(f(x)+xf'(x))dx.$$ Since $f(\pi)=0$, $$B = -\frac{1}{2}\int_{0}^{\pi} x^2f'(x)dx.$$ Likewise, we get that $$A = -\pi\int_{0}^{\pi}xf'(x)dx.$$ However, I don't see how to advance using integration by parts.

Can you give me some advice to find the desired value?

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2 Answers 2

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Lemma: $$ \int_{0}^{\pi} xf(\sin(x))\, dx = \frac{\pi}{2}\int_{0}^{\pi} f(\sin(x)) \, dx$$ Proof: Taking the substitution $u = \pi -x $ gives us $$ \int_{0}^{\pi} xf(\sin(x))\, dx = \int_{\color{blue}{\pi}}^{\color{blue}{0}}(\pi - u)f(\sin(\pi - u)) (\color{blue}{- \, du}) = \int_{0}^{\pi} (\pi - u)f(\sin(\pi - u)) \, du $$ but since $\sin(\pi - u)=\sin(u)$ from the definition of the sine function, we see that $\require{cancel}$ \begin{align} \underbrace{ \int_{0}^{\pi} xf(\sin(x))\, dx }_{\color{purple}{I}} = \int_{0}^{\pi} (\pi - u)f(\sin(u)) \, du &= \pi\int_{0}^{\pi} f(\sin(u)) \, du - \underbrace{\int_{0}^{\pi} uf(\sin(u)) \, du}_{\color{purple}{I}}\\ \implies 2 \color{purple}{I} &= \pi\int_{0}^{\pi} f(\sin(u))\, du\\ \implies \int_{0}^{\pi} xf(\sin(x))\, dx &= I = \frac{\pi}{2}\int_{0}^{\pi} f(\sin(u)) \, du \end{align} Q.E.D.


With the previous lemma the problem becomes simple. Notice that $\frac{\sin(x)}{1+ \sin(x)}$ is indeed $f(x) = \frac{x}{1+x}$ composed with $\sin(x)$, so we can apply the lemma to integral $B$ and get that $$ B = \int_{0}^{\pi} x\underbrace{\frac{\sin(x)}{1+ \sin(x)}}_{f(\sin(x))} \, dx = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin(x)}{1+ \sin(x)} \,dx \tag{1} $$ Now, for $A$ we take the substitution $u = \pi x$. This gives $$ A = \pi^2 \int_{0}^{1}\frac{\sin(\pi x)}{1+ \sin(\pi x)}\, dx = \pi^{\cancel{2}}\int_{\color{blue}{0}}^{\color{blue}{\pi}}\frac{\sin(u)}{1+ \sin(u)}\left( \frac{\color{blue}{du}}{\cancel{\color{blue}{\pi}}}\right)=\pi \int_{0}^{\pi} \frac{\sin(u)}{1+ \sin(u)} \,du \tag{2} $$ And combining equations $(1)$ and $(2)$ we get $$ \frac{A}{B} = \frac{\cancel{\pi} \cancel{\int_{0}^{\pi} \frac{\sin(u)}{1+ \sin(u)} \,du}}{\frac{\cancel{\pi}}{2} \cancel{\int_{0}^{\pi} \frac{\sin(x)}{1+ \sin(x)} \, dx}} = \boxed{2} $$

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    $\begingroup$ +1 for the nice Lemma :). $\endgroup$ Nov 25, 2021 at 22:43
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Using Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$.,

$$B+B=\pi\int\dfrac{\sin x}{1+\sin x}dx$$

Set $x=\pi y\implies dx=\pi dy$ to find $$B+B=A$$

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