How to deal with odd $m$ in integral $\int_{0}^{\frac{\pi}{4}}(\sin^{6}m x+\cos^{6}m x) \ln (1+\tan x) d x $ Latest edit

*

*Thanks to @Quanto for settling down the question by proving the odd one as:$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)$$
By our results for both odd and even multiples $n$ of $x$, we can conclude that

$$
\lim _{n \rightarrow \infty} \displaystyle \int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(nx)+\cos ^{6}(nx)\right] \ln (1+\tan x) d x =\frac{5 \pi\ln 2}{64}
$$


*As asked by @Claude Leibovici for the powers other than 6, I had generalised my result to even powers below as an answer:
$$
I(m,n):=\int_{0}^{\frac{\pi}{4}}\left[\cos ^{2 m}(2 nx)+\sin ^{2 m}(2 n x)\right] \ln (1+\tan x) d x= \frac{\pi \ln 2}{4} \cdot \frac{(2 m-1) ! !}{(2 m) ! !}
$$

In order to evaluate the even case
$$\int_{0}^{\frac{\pi}{4}}\left[\sin^{6}(2 n x)+\cos^{6}(2 nx)\right] \ln (1+\tan x) d x $$
we first simplify
$\displaystyle \begin{aligned}\sin ^{6}(2 n x)+\cos ^{6}(2 n x) =& {\left[\sin ^{2}(2 n x)+\cos ^{2}(2 n x)\right]\left[\sin ^{4}(2 n x)-\sin ^{2}(2 n x) \cos ^{2}(2 n x)\right) } \\&\left.+\cos ^{4}(2 n x)\right] \\=& 1-3 \sin ^{2}(2 n x) \cos ^{2}(2 n x) \\=& 1-\frac{3}{4} \sin ^{2}(4 n x) \\=& 1-\frac{3}{8}(1-\cos 8 n x) \\=& \frac{1}{8}(5+3 \cos (8nx))\end{aligned} \tag*{} $
To get rid of the natural logarithm, a simple substitution transforms the integral into
$\begin{aligned}I &=\frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln (1+\tan x) d x \\& \stackrel{x\mapsto\frac{\pi}{4}-x}{=} \frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x \\&=\frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln \left(\frac{2}{1+\tan x}\right) d x \\&=\frac{1}{8} \ln 2 \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x) )d x-I \\I &=\frac{\ln 2}{16} \int_{0}^{\frac{\pi}{4}}(5+3 \cos 8  n x) d x\\&=\frac{\ln 2}{16}\left[5 x+\frac{3}{8 n} \sin (8 n x)\right]_0^{\frac{\pi}{4} }\\ &=\frac{5 \pi}{64} \ln 2\end{aligned} \tag*{} $
My Question:
How can we deal with the odd one
$$\displaystyle \int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(2 n +1)x+\cos ^{6}(2 n +1)x\right] \ln (1+\tan x) d x  ?$$
Can you help?
 A: Similar to the even case, recognize
$$\sin ^{6}mx+\cos ^{6}mx = \frac58+\frac38 \cos4mx $$ and rewrite the integral as follows
\begin{align}
I_{2n+1}=&\int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(2 n +1)x+\cos ^{6}(2 n +1)x\right] \ln (1+\tan x) d x  \\
=&\> \frac58\int_{0}^{\frac{\pi}{4}}\ln \overset{\frac\pi4-x \to x}{(1+\tan x) } d x 
+ \frac38\int_{0}^{\frac{\pi}{4}}
\cos4(2n+1)x\ln\frac{\sqrt2\cos(\overset{\frac\pi4-x \to x}{\frac\pi4-x})}{\cos x}dx\\
=&\> \frac58\cdot \frac\pi8\ln2- 
\frac34\int_{0}^{\frac{\pi}{4}}
\cos4(2n+1)x\ln \cos x\> \overset{ibp}{dx}\\
=&\> \frac{5\pi}{64}\ln2-\frac3{16(2n+1)}
\int_{0}^{\frac{\pi}{4}}\frac{\sin 4(2n+1)x\sin x}{\cos x}dx\tag1
\end{align}
Derive the integral below recursively
\begin{align}
K_m=&\int_{0}^{\frac{\pi}{4}}\frac{\sin 4m x\sin x}{\cos x}dx
= K_{m-1}+\frac{(-1)^{m-1}}{2m-1}=-\frac\pi4+\sum_{j=0}^{m-1} \frac{(-1)^j}{2j+1}\\
\end{align}
Then, evaluate $K_{2n+1}$ and plug into (1) to obtain
$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)
$$
In contrast, the even case $I_{2n}= \frac{5\pi}{64}\ln2$ is much simpler.
A: Thanks to @Quanto for settling down the question by proving the odd one as:$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)
$$
As asked by @Claude Leibovici for the powers other than 6, I had generalised my result to even powers below:
$$
I(m,n):=\int_{0}^{\frac{\pi}{4}}\left[\cos ^{2 m}(2 nx)+\sin ^{2 m}(2 n x)\right] \ln (1+\tan x) d x= \frac{\pi \ln 2}{4} \cdot \frac{(2 m-1) ! !}{(2 m) ! !}
$$
Proof:
Letting $x\mapsto \frac{\pi}{4}-x $ yields
$$
\begin{aligned}
I(m, n) &=\int_{0}^{\frac{\pi}{4}}\left[\sin ^{2 m}(2 x)+\cos ^{2 m}(2 x)\right) \ln \left(\frac{2}{1+\tan x}\right]d x \\
&=\ln 2 \int_{0}^{\frac{\pi}{4}}\left[\sin^{2m} (2 n x)+\cos ^{2 m}(2 n x) \right] d x-I(m, n)\\ I(m, n)&= \frac{\ln 2}{2} \int_{0}^{\frac{\pi}{4}}\left[\sin^{2m} (2n x)+\cos ^{2 m}(2nx)\right]d x\\& = \frac{\ln 2}{4} \int_{0}^{\frac{\pi}{2}}\left[\sin^{2m} (n x)+\cos ^{2 m}(nx)\right]d x\\&\stackrel{nx\mapsto x}  {=}\frac{\ln 2}{2} \cdot \frac{1}{n} \int_{0}^{\frac{n \pi}{2}} \sin ^{2 m} x d x\\&= \frac{\ln 2}{2} \int_{0}^{\frac{\pi}{2}} \sin ^{2 m} x d x
\end{aligned}
$$
Using Wallis Formula, we can conclude that $$
\boxed{I(m, n)=\frac{\pi \ln 2}{4} \cdot \frac{(2 m-1) ! !}{(2 m) ! !}},
$$
which is independent of $n$.
A: Since @Quanto provided a detailed and very good answer, I shall make the story short.
Working directly the problem of the difference between two consecutive terms
$$I_{2n+1}-I_{2n}=-\frac{3 }{32 } \frac{1 }{2 n+1}\Phi \left(-1,1,\frac{4n+3}{2}\right)$$ where appears the Lerch transcendent function.
We can also write it as
$$I_{2n+1}-I_{2n}=-\frac{3 }{64 }\frac{1 }{2 n+1} \left(H_{n+\frac{1}{4}}-H_{n-\frac{1}{4}}\right)$$ These are very small numbers. Expanded as series
$$I_{2n+1}-I_{2n}=-\frac{3}{256 n^2}\Bigg[1-\frac{1}{n}+\frac{11}{16 n^2}-\frac{3}{8 n^3}+O\left(\frac{1}{n^4}\right) \Bigg]$$
