How many methods are there to find the closed form of $ \int_{0}^{\frac{\pi}{2}} \frac{d x}{(\sin x+\cos x)^{ n}}, \textrm{ where } n\in N $

Question

Latest Edit

Great thanks to @Quanto for his closed form for

$$\int_{0}^{\frac{\pi}{4}} \sec^{2n+1} x \ dx= \frac1{2^{2n}} \binom {2n}n \ln(\sqrt2+1) +\sum_{k=0}^{n-1} \binom {2n}k \frac{(\sqrt2+1)^{2(n-k)} - (\sqrt2-1)^{2(n-k)}}{2^{2n+1}(n-k)}, $$

by which the closed form for integrals with odd powers can be found as

$$\boxed{\quad \int_{0}^{\frac{\pi}{2}} \frac{d x}{(\sin x+\cos x)^{2 n+1}}\\= \frac1{2^{3n-\frac{1}{2} }} \binom {2n}n \ln(\sqrt2+1) +\sum_{k=0}^{n-1} \binom {2n}k \frac{(\sqrt2+1)^{2(n-k)} - (\sqrt2-1)^{2(n-k)}}{2^{3n+\frac{1}{2} }(n-k)} } $$

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Inspired by the post, I firstly try to generalise it to even powers as $$ I_{n}=\int_{0}^{\frac{\pi}{2}} \frac{d x}{(\sin x+\cos x)^{2 n}}, $$

where $n\in N.$

For simplicity, we express the denominator as cosine using compound angle formula. $$ \begin{aligned} I_{n} &=\int_{0}^{\frac{\pi}{2}} \frac{d x}{(\sin x+\cos x)^{2 n}} \\ &=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left[\sqrt{2} \cos \left(\frac{\pi}{4}-x\right)\right]^{2 n}} \\ & \stackrel{x\mapsto\frac{\pi}{2}-x}{=} \frac{1}{2^{n}}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\cos ^{2 n} x}\\ & \stackrel{Symm}{=} \frac{1}{2^{n-1}} \int_{0}^{\frac{\pi}{4}} \sec ^{2 n} x d x \end{aligned} $$

Letting $t= \tan x$ yields $$ \begin{aligned} I_{n} &= \frac{1}{2^{n-1}}\int_{0}^{1}\left(1+t^{2}\right)^{n-1} d t \\ &= \frac{1}{2^{n-1}} \sum_{k=0}^{n-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) \int_{0}^{1} t^{2 k} d t \\ &=\boxed{\frac{1}{2^{n-1}} \sum_{k=0}^{n-1}\left(\begin{array}{l} n-1 \\ \quad k \end{array}\right) \frac{1}{2 k+1}} \end{aligned} $$

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Examples: $$ I_{2} =\frac{1}{2}\left(1+\frac{1}{3}\right)=\frac{2}{3},I_{5}=\frac{83}{315}; I_{10}=\frac{26206}{230945}; $$

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However, it is harder to find the integral of odd powers. $$ J_{n}=\int_{0}^{\frac{\pi}{2}} \frac{d x}{(\sin x+\cos x)^{2 n+1}}= \frac{1}{2^{n-\frac{1}{2} }} \int_{0}^{\frac{\pi}{4}} \sec ^{2 n+1} x d x $$ where $n=0, 1, 2,…$

I had found a reduction formula for it as $$ \boxed{J_{n+1}=\frac{1}{2 n}+\frac{2 n-1}{4 n} J_{n}} $$

via the reduction formula for $$ \int_{0}^{\frac{\pi}{4}} \sec ^{2 n+1} x d x=\frac{1}{2 n}\left[2^{n-\frac{1}{2}}+(2 n-1) \int_{0}^{\frac{\pi}{4}} \sec ^{2 n-1} x d x\right] $$

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Examples

$$ \begin{aligned} J_{0} &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \sec x d x \\ &=\sqrt{2}\left|\ln \left|\sec x+\tan x\right|\right]_{0} ^{\frac{\pi}{4}}\\ &=\sqrt{2} \ln (\sqrt{2}+1) \end{aligned} $$

\begin{align} \begin{aligned} J_{1} &=\frac{1}{2}+\frac{1}{4} \cdot \sqrt{2} \ln (\sqrt{2}+1) \\ &=\frac{1}{2}+\frac{1}{2 \sqrt{2}} \ln (\sqrt{2}+1) \end{aligned} \end{align}

$$ \begin{aligned} J_{2} &=\frac{1}{4}+\frac{3}{8}\left[\frac{1}{2}+\frac{1}{2 \sqrt{2}} \ln (\sqrt{2}+1)\right] \\ &=\frac{7}{16}+\frac{3}{16 \sqrt{2}} \ln (\sqrt{2}+1) \end{aligned} $$

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Although I can find the integral with odd powers one by one, I still can’t find its closed form. Your help will be highly appreciated.

Answer 1

For the odd case, substitute $\sec x= \cosh t$ to integrate \begin{align} &\int_{0}^{\frac{\pi}{4}} \sec^{2n+1} x \ d x =\int_0^{\ln(\sqrt2+1)}\cosh^{2n}t\ dt\\ =& \ \frac1{2^{2n}}\int_0^{\ln(\sqrt2+1)}\bigg[ \binom {2n}n+2 \sum_{k=0}^{n-1} \binom {2n}k \cosh2(n-k)t\bigg]\ dt\\ = &\ \frac1{2^{2n}} \binom {2n}n \ln(\sqrt2+1) +\sum_{k=0}^{n-1} \binom {2n}k \frac{(\sqrt2+1)^{2(n-k)} - (\sqrt2-1)^{2(n-k)}}{2^{2n+1}(n-k)} \\ \end{align}

Answer 2

Maple solves your recurrence as: $$ J_n= {\frac {\Gamma \left( n+{ \frac{3}{2}} \right)}{4\,\sqrt {\pi}\Gamma \left( n+2 \right) } \left( \sqrt {\pi}\sum _{k=0}^{n}{{2}^{k+1} \frac{\Gamma \left( k+1 \right)} { \Gamma \left( k+{\frac{3}{2}} \right)}} {2}^{-n}+{2}^{1-n}J_1 \right) } $$