TITLE: Calculate the following integral $\int_0^{\pi/2} \frac{\sin^m x\,\mathrm{d}x}{\sin x + \cos x}$, $m=2k-1$
QUESTION [14 upvotes]: At the moment I am studing the following integral

$$K(m,n)= \int_0^{\pi/2} \frac{\sin^m x\,\mathrm{d}x}{\sin^nx + \cos^nx}.$$

For integers $m$,$n$. The question regarding both $K(1,1)$ and
$K(n,n)$ has been asked countless of times before on the site. With the usual solution of
$x \mapsto \pi/2 - x$. I have tried to study the case $K(2k-1,1)$ for integers $k$.
It seems this integral does not take any elementary form, but can it be expressed in a easier fashion? 
Beneath is my work that proved more or less useless.. Assume that $m$ can be written on the form $2k-1$ where $k$ is some positive integer. Then we have from the binomial expansion that 
$$
\frac{a^n + b^n}{a+b} = a^{n-1} - a^{n-2}b + a^{n-3}b^2 + \cdots - a b^{n-2} + b^{n-1}
$$
for odd $n$. By using the usual substitution one obtains that 
$$
    K(m,1) = \frac{1}{2} \int_0^{\pi/2} \frac{\sin^mx + \cos^mx}{\cos x + \sin x} \mathrm{d}x
           = \frac{1}{2} \int_0^{\pi/2} \left( \sum_{k=1}^m \cos^{m-k}(x)\sin^{k-1}(x)\right)\mathrm{d}x
$$
Since the integral is convergent for any positive $k$, both $\sin x$ and $\cos x$ live in $L^2$ we can switch the limits. By using the standard definition of the beta integral, the integral turns into 
$$
    K(m,1) = \frac{1}{4} \sum_{k=1}^m \operatorname{B}(x,y)
$$
With $2x-1=m-k$ and $2y-1=k-1$. But this does not seem any easier.. 
By using the symmetri $\operatorname{B}(y,x)=\operatorname{B}(x,y)$, 
and inputing $m = 2p-1$, the integral can also be written as
$$
    K(2p-1,1) = \frac{1}{4}\operatorname{B}\left(p^*,p^*\right) 
              + \frac{1}{2}\sum_{k=1}^{p-1} \operatorname{B}(p-k^*,k^*)
$$
With $p^*=p/2$ and $k^*=k/2$ to make the notation slightly better. 
Are there any other perhaps more elementary way to find an expression for the integral? Is it possible to simplify this expression more?

REPLY [2 votes]: I intend to develop some of N3buchadnezzar's ideas and correct some changes of sign which he has lost. The result will generalize the calculations that user17762 and  Arbias Hashani have started.
From the definition of the $\Gamma$ function and a change of variable I get that
\begin{equation*}
\Gamma(\alpha) = \int_{0}^{\infty}x^{\alpha -1}e^{-x}\, dx = 2 \int_{0}^{\infty}x^{2\alpha -1}e^{-x^{2}}\, dx
\end{equation*}
where $\alpha>0$.
Let $D$ be a notation for the first quadrant. If I change to polar coordinates I get
\begin{gather*}
\Gamma(\alpha)\Gamma(\beta) = 4\iint_{D}x^{2\alpha-1}y^{2\beta-1}e^{-x^{2}-y^{2}}\, dxdy =\\ 2\int_{0}^{\pi^/2}\cos^{2\alpha-1}(\theta)\sin^{2\beta-1}(\theta)\, d\theta \cdot 2\int_{0}^{\infty}r^{2a+2\beta-1}e^{-r^{2}}\, dr = \\2\Gamma(\alpha+\beta)\cdot \int_{0}^{\pi^/2}\cos^{2\alpha-1}(\theta)\sin^{2\beta-1}(\theta)\, d\theta.
\end{gather*}
Consequently
\begin{equation*}
\int_{0}^{\pi^/2}\cos^{2\alpha-1}(\theta)\sin^{2\beta-1}(\theta)\, d\theta = \dfrac{\Gamma(\alpha)\Gamma(\beta)}{2\Gamma(\alpha+\beta)} = \dfrac{1}{2}B(\alpha,\beta)\quad\text{ if } \alpha>0, \beta >0 \tag{1}
\end{equation*}
where $B$ is the Beta function. However I will not use it.(cf. N3buchadnezzar.)
Put
\begin{equation*}
I_k = \int_{0}^{\pi/2}\dfrac{\sin^{2k-1}(x)}{\sin(x) + \cos(x)}\, dx , \quad k\in \mathbb{N}. 
\end{equation*}
By symmetry 
\begin{equation*}
I_k = \dfrac{1}{2}\int_{0}^{\pi/2}\dfrac{\cos^{2k-1}(x) +\sin^{2k-1}(x)}{\cos(x) + \sin(x)}\, dx.
\end{equation*}
In the formula
\begin{equation*}
\dfrac{a^{2k-1} + b^{2k-1}}{a+b} =\sum_{j=1}^{k}a^{2k-2j}b^{2j-2} - \sum_{j=1}^{k-1}a^{2k-2j-1}b^{2j-1}. \tag{2}
\end{equation*}
I have collected all positive terms and all negative terms into two separate sums. (cf. N3buchadnezzar.)
According to (1) and (2)  I get that
\begin{gather*}
I_k = \dfrac{1}{2}\sum_{j=1}^{k}\int_{0}^{\pi/2}\cos^{2k-2j}(x)\sin^{2j-2}(x)\, dx - \dfrac{1}{2}\sum_{j=1}^{k-1}\int_{0}^{\pi/2}\cos^{2k-2j-1}(x)\sin^{2j-1}(x)\, dx = \notag\\[2ex]
\dfrac{1}{4}\sum_{j=1}^{k}\dfrac{\Gamma(k-j + \frac{1}{2})\Gamma(j-\frac{1}{2})}{\Gamma(k)}-\dfrac{1}{4}\sum_{j=1}^{k-1}\dfrac{\Gamma(k-j)\Gamma(j)}{\Gamma(k)} = \notag\\[2ex]
\dfrac{1}{4(k-1)!}\left(\sum_{j=1}^{k}\Gamma(k-j + \frac{1}{2})\Gamma(j-\frac{1}{2}) - \sum_{j=1}^{k-1}(k-j-1)!(j-1)!\right). \tag{3}
\end{gather*}
But
\begin{equation*}
\Gamma(n+\dfrac{1}{2}) = \dfrac{(2n-1)!!}{2^{n}}\sqrt{\pi}= \dfrac{(2n)!}{4^{n}n!}\sqrt{\pi}\quad \text{ if } n\in \mathbb{N}.
\end{equation*}
Thus
\begin{equation*}
\Gamma(k-j + \frac{1}{2})\Gamma(j-\frac{1}{2}) = \dfrac{(2k-2j)!}{4^{k-j}(k-j)!}\sqrt{\pi}\cdot\dfrac{(2j-2)!}{4^{j-1}(j-1)!}\sqrt{\pi} = \dfrac{(2k-2j)!(2j-2)!}{4^{k-1}(k-j)!(j-1)!}\pi.
\end{equation*}
Now I can write (3) as
\begin{equation*}
I_k = \dfrac{\pi}{4^{k}(k-1)!}\sum_{j=1}^{k}\dfrac{(2k-2j)!(2j-2)!}{(k-j)!(j-1)!} - \dfrac{1}{4(k-1)!}\sum_{j=1}^{k-1}(k-j-1)!(j-1)!.
\end{equation*}
Remark. Calculations by user17762 indicate that an even $m$ will give the integral 
\begin{equation*}
\int_{0}^{\pi/2}\dfrac{\sin^{m}x}{\sin x + \cos x}\, dx
\end{equation*}
another character than an odd $m$.
With $m=2k$ I can split the integral into two integrals. The first one of them is easily calculated e.g. via 
$ t = \tan\dfrac{x}{2}$. The second integral is a little more intricate.
\begin{gather*}
\int_{0}^{\pi/2}\dfrac{\sin^{2k}x}{\sin x + \cos x}\, dx = 2^{-k}\int_{0}^{\pi/2}\dfrac{1}{\cos x + \sin x}\, dx + \int_{0}^{\pi/2}\dfrac{\sin^{2k}x - 2^{-k}}{\cos x + \sin x}\, dx=\\[2ex]
2^{-k}\sqrt{2}\ln(\sqrt{2}+1) + 2^{-k}k!\sum_{j=1}^{\lfloor{k/2}\rfloor}\sum_{l=0}^{2j-1}\dfrac{(-2)^{l}}{j\cdot(2l+1)\cdot l!(k-2j)!(2j-l-1)!}.
\end{gather*}
To verify the double sum I have to do some calculations.
At first I rewrite the numerator in the integrand.
\begin{equation*}
\sin^{2k}x - 2^{-k} = 2^{-k}(1-\cos 2x)^{k}-2^{-k} = 2^{-k}\sum_{j=1}^{k}\binom{k}{j}(-1)^{j}\cos^{j}(2x).
\end{equation*}
Since $\cos 2x = \cos^{2}x - \sin^{2}x$ the integrand can be rewritten as
\begin{equation*}
\dfrac{\sin^{2k}x - 2^{-k}}{\cos x + \sin x} = 2^{-k}\sum_{j=1}^{k}\binom{k}{j}(-1)^{j}\cos^{j-1}(2x)(\cos x - \sin x).
\end{equation*}
The change of variable $x = \dfrac{\pi}{2}-t$ proves that
\begin{equation*}
\int_{0}^{\pi/2}\cos^{j-1}(2x)(\cos x - \sin x)\, dx = 0
\end{equation*}
if $j$ is odd and also that
\begin{equation*}
\int_{0}^{\pi/2}\cos^{j-1}(2x)\cos x\, dx = -\int_{0}^{\pi/2}\cos^{j-1}(2x)\sin x\, dx.
\end{equation*}
if $j$ is even. Consequently
\begin{equation*}
\int_{0}^{\pi/2}\dfrac{\sin^{2k}x - 2^{-k}}{\cos x + \sin x}\, dx = 2^{1-k}\sum_{j=1}^{\lfloor{k/2}\rfloor}\binom{k}{2j}\int_{0}^{\pi/2}\cos^{2j-1}(2x)\cos x\, dx.\tag {4}
\end{equation*}
Futhermore
\begin{equation*}
\cos^{2j-1}(2x)\cos x = (1-2\sin^{2}x)^{2j-1}\cos x = \sum_{l=0}^{2j-1}\binom{2j-1}{l}(-2)^{l}\sin^{2l}(x)\cos x.
\end{equation*}
Thus
\begin{equation*}
\int_{0}^{\pi/2}\cos^{2j-1}(2x)\cos x\, dx = \sum_{l=0}^{2j-1}\binom{2j-1}{l}(-2)^{l}\dfrac{1}{2l+1}.\tag{5}
\end{equation*}
Since
\begin{equation*}
\binom{k}{2j}\binom{2j-1}{l}\dfrac{1}{2l+1} = \dfrac{k!}{2j\cdot(2l+1)\cdot l!(k-2j)!(2j-l-1)!}
\end{equation*}
I can combine (4) and (5) to
\begin{equation*}
\int_{0}^{\pi/2}\dfrac{\sin^{2k}x - 2^{-k}}{\cos x + \sin x}\, dx = 2^{-k}k!\sum_{j=1}^{\lfloor{k/2}\rfloor}\sum_{l=0}^{2j-1}\dfrac{(-2)^{l}}{j\cdot(2l+1)\cdot l!(k-2j)!(2j-l-1)!}.
\end{equation*}