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Given the integral $(1)$

$$\int_{0}^{\pi}{\sin^5 x\over 1+\cos^3 x}\mathrm dx=\ln 3\tag1$$

This integral seems to be quite tough, I can't even know where to start. I have try substitution $u=1+\cos^3 x$ and $u=\cos^3 x$, can seem to remove the x variables.

We have the following identities

$$\sin^5a={5\over 8}\sin a-{5\over 16}\sin 3a+{1\over 16}\sin 5a\tag2$$

$$\cos^3 a={3\over 4}\cos a+{1\over 4}\cos 3a\tag3$$

We may write $(1)$ as

$$\int_{0}^{\pi}{\sin x(1-\cos^2 x)^2\over 1+\cos x(1-\sin ^2 x)}\mathrm dx\tag4$$

$u=1-\sin^2 x\implies \mathrm du=-2\sin x\cos x\mathrm dx$ then $(4)$ becomes

$${1\over 2}\int_{0}^{1}{(1-u)^2\over \sqrt{u}+u^2}\mathrm du\tag5$$

$u=v^2\implies du=2vdv$ then $(5)$ becomes

$$\int_{0}^{1}{(1-v^2)^2\over v+v^4}\cdot{v\mathrm dv}\tag6$$ Simplify to

$$\int_{0}^{1}{(1-v)(1-v^2)\over 1-v+v^2}\mathrm dv\tag7$$

I guess using partial decomposition of fraction and integrate term by term, seem lengthy process

How else can we prove $(1)$?

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    Note: The substitution $v=\cos x$ gets you from beginning to end of your post more quickly. The key to this is the odd power of $\sin x$ in the original numerator. – Mark Bennet Mar 22 '17 at 18:56
  • Weierstrass substitution followed by an application of the residue theorem seems promising – tired Mar 22 '17 at 18:57
  • The integral equals

    $$ I=2\int_0^1\frac{(\cos(x)-1)^2 d(\cos(x))}{1+\cos^3(x)} $$

     – tired Mar 22 '17 at 19:00
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    $$ \frac{(1-v)(1-v^2)}{1-v+v^2} = v - \frac{2v-1}{1-v+v^2}$$ Note that $\dfrac{d}{dv} (1-v+v^2) = 2v-1$. – Robert Israel Mar 22 '17 at 19:06
  • Just to note (apologies for not spotting it) you need to take care about the limits of the integral. – Mark Bennet Mar 22 '17 at 19:53

2 Answers2

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  1. Splitting $\sin^5 x=\sin^4x\cdot\sin x=(1-\cos^2x)^2\cdot\sin x$ and making the substitution $v=\cos x$: $$ \int_{-1}^1\frac{(1-v^2)^2}{1+v^3}\,dv. $$
  2. Cancelling $v+1$ and doing the polynomial division: $$ \int_{-1}^1\left(v-\frac{2v-1}{v^2-v+1}\right)\,dv. $$
  3. The antiderivative is (the easy chain rule in the fraction): $$ \frac{v^2}{2}-\ln(v^2-v+1)+C. $$
A.Γ.
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The given integral equals $$ \int_{0}^{\pi/2}\frac{\sin^5(x)}{1+\cos^3(x)}\,dx +\int_{0}^{\pi/2}\frac{\sin^5(x)}{1-\cos^3(x)} = 2\int_{0}^{\pi/2}\frac{\sin^5(x)}{1-\cos^6(x)}\,dx $$ or just $$ 2\int_{0}^{1}\frac{(1-u^2)^2}{1-u^6}\,du = 2\sum_{k\geq 0}\left(\frac{1}{6k+1}-\frac{2}{6k+3}+\frac{1}{6k+5}\right)$$ that equals $$ \frac{1}{3}\left[2\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{6}\right)-\psi\left(\frac{5}{6}\right)\right] $$ or $\log(3)$, by the reflection and duplication formulas for the digamma function.

Jack D'Aurizio
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