How to tackle the integral $\int_{0}^{\pi} \frac{3 \pi x^{2}-2 x^{3}}{(1+\sin x)^{n}}dx, \textrm{ where } n \in N $

Question

Inspired by an integral in my post, I want to generalize the result by replacing the power 2 in the denominator by $n.$

Letting $x \mapsto\pi-x$ yields $$ I_n:=\int_{0}^{\pi} \frac{3 \pi x^{2}-2 x^{3}}{(1+\sin x)^{n}} d x=\int_{0}^{\pi} \frac{2 x^{3}-3 \pi x^{2}+\pi^{3}}{(1+\sin x)^{n}} d x=-I+\pi^{3} \int_{0}^{\pi} \frac{d x}{(1+\sin x)^{n}} $$

Rearranging gives $$ I_{n}=\frac{\pi^{3}}{2} \underbrace{\int_{0}^{\pi} \frac{d x}{(1+\sin x)^{n}}}_{K_n} $$

$$ \begin{aligned} K_{n} & \stackrel{2 y=\frac{\pi}{2}- x}{=} 2 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d y}{(1+\cos 2 y)^{n}} \\ &=2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d y}{\left(2 \cos ^{2} y\right)^{n}}\\&= \frac{1}{2^{n-2}} \underbrace{\int_{0}^{\frac{\pi}{4}} \sec ^{2 n} y d y}_{J_n} \end{aligned} $$

Now we need a reduction formula for $J_n$, for any natural number $n$, $$ \begin{aligned} J_{n} &=\int_{0}^{\frac{\pi}{4}} \sec ^{2 n} y d y\\ &=\int_{0}^{\frac{\pi}{4}} \sec ^{2 n-2} y d(\tan y) \\ &=\left[\sec ^{2 n-2} y \tan y\right]_{0}^{\frac{\pi}{4}}-(2 n-2) \int_{0}^{\frac{\pi}{4}} \tan ^{2} y \sec ^{2 n-3} y \sec y dy\\ &=2^{n-1}-(2 n-2) \int_{0}^{\frac{\pi}{4}}\left(\sec ^{2} y-1\right) \sec ^{2 n-2} y d y \\ &=2^{n-1}-(2 n-2) \int_{0}^{\frac{\pi}{4}} \sec ^{2 n} y d y+(2 n-2) J_{n-1} \end{aligned} $$

Rearranging yields the reduction formula $$ J_{n}=\frac{1}{2 n-1}\left[2^{n-1}+(2 n-2) J_{n-1}\right] \tag*{(1)} $$

Hence $$ I_{n}=\frac{\pi^{3}}{2} K_{n}=\frac{\pi^{3}}{2} \cdot \frac{1}{2^{n-2}} J_{n}=\frac{\pi^{3}}{2^{n-1}} J n\tag*{(2)} $$

Back to the reduction formula for our integral $I_n$, combining (1) and (2) gives $$ \boxed{I_{n}=\frac{1}{2 n-1}\left[\pi^{3}+(n-1) I_{n-1}\right]} $$

For examples,

$$ \displaystyle \begin{array}{l} I_{1}=\frac{1}{1}\left(\pi^{3}+0\cdot I_{0}\right)=\pi^{3} \\ I_{2}=\frac{1}{3}\left(\pi^{3}+1 \cdot I_{1}\right)=\frac{2 \pi^{3}}{3} \\ I_{3}=\frac{1}{5}\left(\pi^{3}+2 \cdot I_{2}\right)=\frac{7 \pi^{3}}{15} \\ I_{4}=\frac{1}{7}\left(\pi^{3}+3 \cdot I_{3}\right)=\frac{12 \pi^{3}}{35}\\ \qquad\qquad \vdots \end{array} $$

My Question

Although we can find our integral one by one by the reduction formula, it is tedious and unsatisfactory. Is there any closed form for it?

Latest Edit

Helped by Mr Quanto, we got a closed form for the integral $$ \boxed{I_{n}=\frac{\pi^{3}}{2^{n-1}} \sum_{k=0}^{n-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) \frac{1}{2 k+1}} $$

which is obtained by letting $t=\tan y$ \begin{aligned} J_{n} &=\int_{0}^{\frac{\pi}{4}} \sec ^{2 n} y d y \\ &=\int_{0}^{1}\left(1+t^{2}\right)^{n-1} d t \\ &=\sum_{k=0}^{n-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) \int_{0}^{1} t^{2 k} d t \\ &=\sum_{k=0}^{n-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) \frac{1}{2 k+1} \end{aligned}

For example $$ \begin{aligned} I_{5} &=\frac{\pi^{3}}{2^{4}}\left[\left(\begin{array}{l} 4 \\ 0 \end{array}\right)+\left(\begin{array}{l} 4 \\ 1 \end{array}\right) \frac{1}{3}+\left(\begin{array}{l} 4 \\ 2 \end{array}\right) \frac{1}{5}+\left(\begin{array}{l} 4 \\ 3 \end{array}\right) \frac{1}{7}+\left(\begin{array}{l} 4 \\ 4 \end{array}\right) \frac{1}{9}\right] =\frac{83 \pi^{3}}{315}\approx{8.16991}, \end{aligned} $$ which is checked by Wolframalpha

Answer 1

Let $t=\tan y$\begin{aligned} J_{n} &=\int_{0}^{\frac{\pi}{4}} \sec ^{2 n} y \>d y= \int_0^1 (t^2+1)^{n-1}dt= \sum_{k=0}^{n-1}\frac{\binom{n-1}k}{2(n-k)-1} \end{aligned}

Answer 2

Even if it is messy, without any change of variable, there is an explicit antiderivative (have a look here) in terms of hypergeometric functions.

Using the bounds and simplifying as much as possible leads to $$I_n=\int_{0}^{\pi} \frac{3 \pi x^{2}-2 x^{3}}{[1+\sin (x)]^{n}}\,dx=\frac{\pi ^3}{2^{n-1} } \, _2F_1\left(\frac{1}{2},1-n;\frac{3}{2};-1\right)$$ which gives the reduction formula writen in the post as well as @Quanto's result.