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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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 \(x^5\).

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

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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Great solution. Thanks.

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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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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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Hint: Factorize the denominator into two irreducible quadratic equations.

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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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@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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@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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Can you please solve it for me?

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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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