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Notice that $\frac3{25r^2+25r+4} = \frac{12}{25} \cdot \frac1{(2r+1)^2 - \left (2\cdot \frac3{10} \right)^2}$. Thus $x=\frac3{10}$ for $(\star \star)$.

Hence, the integral in question equals to $\dfrac{12}{25} \cdot \dfrac{\pi}{8\cdot 3/10} \tan\left(\pi \cdot \dfrac3{10}\right) = \dfrac\pi5 \tan\left(\dfrac3{10}\pi\right)$.

Now, we just need to evaluate $\tan\left(\frac3{10}\pi\right)$. Apply the identity $\tan(x) = \sqrt{\frac{1-\cos(2x)}{1+\cos(2x)}}$ for $x = \frac3{10}\pi$.

This means we need to find what is the value of $\cos\left(\frac35\pi \right)$. Let $y$ denote this value. Because $\cos\left(\frac{1}5\pi\right) = -\cos\left( 4\times \frac15\pi \right)$. Apply the double angle formula twice yields $\cos\left( \frac15\pi\right) = \frac{1+\sqrt5}4$. Apply the triple angle formula to get $y = \frac{1-\sqrt5}4$.

Substitution yields the answer of $\displaystyle \pi \sqrt{\dfrac{5+2\sqrt5}{125}} \approx 0.8648 \ \square$

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## Comments

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TopNewestMultiply top and bottom by $-(x-1)$, we have $\displaystyle \int_0^1 \frac{1-x^3}{1-x^5} \, dx$.

Notice that the integrand can be written as a sum of a convergent geometric series with common ratio $x^5$.

$\begin{aligned} \displaystyle \int_0^1 \frac{1-x^3}{1-x^5} \, dx &=& \displaystyle \int_0^1 \displaystyle \sum_{r=0}^\infty (1-x^3) x^{5r} \, dx \\ \displaystyle &=& \sum_{r=0}^\infty \int_0^1 (x^{5r} - x^{3+5r} ) \, dx \\ \displaystyle &=& \sum_{r=0}^\infty \left( \frac1{5r+1} - \frac1{5r+4}\right ) \\ \displaystyle &=& \sum_{r=0}^\infty \frac3{25r^2+25r+4} \qquad (\star) \\ \end{aligned}$

We consider the logarithmic differentiation of the Weiestrass Product:

$\displaystyle \cos(\pi x) = \prod_{n=0}^\infty \left( 1 - \frac{4x^2}{(2n+1)^2} \right)$

to get

$\displaystyle \sum_{n=0}^\infty \frac1{(2n+1)^2 - (2x)^2} = \frac{\pi}{8x} \tan(\pi x ) \qquad (\star \star)$

Notice that $\frac3{25r^2+25r+4} = \frac{12}{25} \cdot \frac1{(2r+1)^2 - \left (2\cdot \frac3{10} \right)^2}$. Thus $x=\frac3{10}$ for $(\star \star)$.

Hence, the integral in question equals to $\dfrac{12}{25} \cdot \dfrac{\pi}{8\cdot 3/10} \tan\left(\pi \cdot \dfrac3{10}\right) = \dfrac\pi5 \tan\left(\dfrac3{10}\pi\right)$.

Now, we just need to evaluate $\tan\left(\frac3{10}\pi\right)$. Apply the identity $\tan(x) = \sqrt{\frac{1-\cos(2x)}{1+\cos(2x)}}$ for $x = \frac3{10}\pi$.

This means we need to find what is the value of $\cos\left(\frac35\pi \right)$. Let $y$ denote this value. Because $\cos\left(\frac{1}5\pi\right) = -\cos\left( 4\times \frac15\pi \right)$. Apply the double angle formula twice yields $\cos\left( \frac15\pi\right) = \frac{1+\sqrt5}4$. Apply the triple angle formula to get $y = \frac{1-\sqrt5}4$.

Substitution yields the answer of $\displaystyle \pi \sqrt{\dfrac{5+2\sqrt5}{125}} \approx 0.8648 \ \square$

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Is it correct to switch the integration and summation?

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Yes, apply Fubini's (or Tonelli's) theorem.

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Great solution. Thanks.

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What makes you think that it has an exact form?

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This question was given to me in my classes and was also told that answer was in terms of pi.

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@Shivam Jadhav divide the denominator by the numerator. Then u can split it up into partial fractions. Then u can proceed it using normal methods

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Wait what? How is that possible? Doesn't it complicate things?

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yes it actually complicates it.

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Can you please solve it for me?

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@Calvin Lin @Tanishq Varshney @Sandeep Bhardwaj

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Check out the Integration of Rational Functions that @Pranshu Gaba and @Vishnuram Leonardodavinci have been working on.

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Answer is $\pi \sqrt{\dfrac{5+2\sqrt5}{125}} \approx 0.8648$. I will post the solution shortly.

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Hint: Factorize the denominator into two irreducible quadratic equations.

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