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How to solve $\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta $?

I have tried substitution of cosine and sine, dividing and multiplying by Sine and cosine, converting everything to tan. It can probably be solved using beta/gamma function but I can't figure it out.

user541396
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  • You can get everything to a $6$-th degree polynomial in $\cos \theta$, and then integrate each term using integration by parts. It will be messy, but it will work. – Joshua Wang Feb 01 '21 at 14:13
  • See here for the last part. https://math.stackexchange.com/questions/193435/prove-int-cosn-x-dx-frac1n-cosn-1x-sin-x-fracn-1n-int-cos – Joshua Wang Feb 01 '21 at 14:14
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    I would use $\int_0^\pi=\frac12\int_{-\pi}^\pi$ (in our case) and express everything in terms of $e^{i\theta}$ (not even expanding the integrand in full detail - only its constant term is needed; the exponentials vanish after integration). – metamorphy Feb 01 '21 at 14:14
  • @metamorphy I understand taking the real part out of $e^(i\theta) - isin\theta $ but after that ? What you mean by not expanding? – user541396 Feb 01 '21 at 18:09
  • @user541396: posted it as an answer. – metamorphy Feb 01 '21 at 19:23

4 Answers4

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Let $\theta =2t $ and use the identities $1+\cos2t=2 \cos^2 t,\> \sin2t = 2\sin t \cos t$ to rewrite the integral as $$I=\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4d\theta = 128\int_0^{\pi/2} (\cos^{10}t - \cos^{12} t )dt$$ Then, apply the recursive result \begin{align} K_n= \int_0^{\pi/2} \cos^ntdt = \frac{n-1}n K_{n-2},\>\>\>\>\>K_0=\frac\pi2 \end{align} to arrive at

$$I=128 (K_{10}- K_{12} )=\frac{32}3K_{10}=\frac{32}3\frac9{10}\frac78\frac56\frac34\frac12K_0=\frac{21}{16}\pi$$

Quanto
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There is another solution using the "reverse" of multiple angle formulae. In your case $$(1-\cos^2(t))(1+\cos(t))^4=$$ $$\frac{3 \cos (t)}{2}-\frac{15 \cos (2 t)}{32}-\frac{5\cos (3 t)}{4} -\frac{13\cos (4 t)}{16} -\frac{\cos (5 t)}{4} -\frac{\cos (6 t)}{32} +\frac{21}{16}$$

Have a look here.

metamorphy
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Claude Leibovici
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The integral is equal to $$\frac12\int_{-\pi}^\pi\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^2\left(1+\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^4d\theta=-\frac{1}{128}\int_{-\pi}^\pi e^{-6i\theta}(e^{2i\theta}-1)^2(e^{i\theta}+1)^8\,d\theta,$$ and if we imagine the integrand written as a linear combination of exponentials, then all the exponentials vanish after the integration, and only its constant term "survives". It is equal to $$[z^6](z^2-1)^2(z+1)^8=\binom{8}{2}-2\binom{8}{4}+\binom{8}{6}=28-2\cdot70+28=-84,$$ so the integral is equal to $\dfrac{84}{128}\cdot2\pi=\dfrac{21}{16}\pi$.

metamorphy
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  • Sorry but what does " if we imagine the integrand written as a linear combination of exponentials, then all the exponentials vanish after the integration, and only its constant term "survives"." mean? I don't follow. Can you please expand? – user541396 Feb 02 '21 at 12:31
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    I don't want to expand ;) Joking. A straightforward way would be to open all the parentheses (that is, expand all the powers/products) in the integrand. You would get a long expression which is a linear combination of $1$, $e^{i\theta}$, $e^{-i\theta}$, $e^{2i\theta}$, etc. (similar to the one @Claude has). And we need only the coefficient of $1$ in this linear combination, since $\int_{-\pi}^\pi e^{in\theta},d\theta=0$ for any integer $n\neq 0$. – metamorphy Feb 02 '21 at 12:41
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Idea: use $\sin^2\theta = 1- \cos^2 \theta$ and work out the polynomial in $\cos \theta$, and apply techniques from here, e.g., or here on this site in a more modern look.

Henno Brandsma
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