Latest Edit
The closed form for $$I_n=
\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{d x}{\sin ^{2n} x+\cos ^{2n} x}
$$
is
$$
\boxed{I_{n}=\frac{\pi^2}{8 n} \sum_{k=0}^{n-1}\left(\begin{array}{c}
n-1 \\
k
\end{array}\right) \csc \frac{(2 k+1) \pi}{2 n}}
$$
Proof:
Letting $t\mapsto \tan x$ yields $$
\begin{aligned}
\int_{0}^{\infty} \frac{\left(1+t^{2}\right)^{n-1}}{t^{2 n}+1}dt &=\sum_{k=0}^{n-1}\left(\begin{array}{c}
n-1 \\
k
\end{array}\right) \int_{0}^{\infty} \frac{t^{2 k}}{t^{2 n}+1} d t .
\end{aligned}
$$
By my post, $$\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x=\frac{\pi}{m} \csc \frac{(r+1) \pi}{m},$$
We can now get its closed form: $$
\boxed{I_{n}=\frac{\pi^2}{8 n} \sum_{k=0}^{n-1}\left(\begin{array}{c}
n-1 \\
k
\end{array}\right) \csc \frac{(2 k+1) \pi}{2 n}}
$$
Original version
As the integral is not so difficult for $n=1,2,3$, I just show how to find the exact value of the integral when $n=4.$
$$
I:=\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin ^{8} x+\cos ^{8} x} d x
,$$
I changed the integral, as usual, by letting $x\mapsto \frac{\pi}{2} -x$, $$I=
\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{d x}{\sin ^{8} x+\cos ^{8} x}
$$
Then multiplying both numerator and denominator by $\sec^8x$ and letting $t=\tan x $ yields $$
I=\frac{\pi}{4} \int_{0}^{\infty} \frac{\left(1+t^{2}\right)^{3}}{t^{8}+1} d t
$$
I was then stuck by the powers and start to think how to reduce them. Thinking for couple of days, I found a way to solve it with partial fractions only. Now I am going to share it with you.
Observing that $$
\int_{0}^{\infty} \frac{d t}{t^{8}+1}\stackrel{t\mapsto\frac{1}{t}}{=}
\int_{0}^{\infty} \frac{t^{6}}{t^{8}+1} d t$$ and
$$
\int_{0}^{\infty} \frac{t^2d t}{t^{8}+1}\stackrel{t\mapsto\frac{1}{t}}{=}
\int_{0}^{\infty} \frac{t^{4}}{t^{8}+1} d t,$$
we can reduce the power of the numerator to 2 that $$
I=\frac{\pi}{2} \underbrace{\int_{0}^{\infty} \frac{1+3 t^{2}}{t^{8}+1} d t}_{J}
$$
To handle the power 8 in the denominator, we resolve the integrand into partial fractions. $$
J=\frac{1}{2 \sqrt{2}} \left[\underbrace{\int_{0}^{\infty} \frac{t^{2}-(3-\sqrt{2})}{t^{4}+\sqrt{2} t^{2}+1} d t}_{K}-\underbrace{\int_{0}^{\infty} \frac{t^{2}-(3+\sqrt{2})}{t^{4}-\sqrt{2} t^{2}+1} d t}_{L} \right]
$$
To deal with $K$ and $L$, we play a little trick. $$
\begin{aligned}
K &=\int_{0}^{\infty} \frac{1-\frac{3-\sqrt{2}}{t^{2}}}{t^{2}+\frac{1}{t^{2}}+\sqrt{2}} d t \\
&=\int_{0}^{\infty} \frac{\frac{\sqrt{2}-2}{2}\left(1+\frac{1}{t^{2}}\right)+\frac{4-\sqrt{2}}{2}\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}+\sqrt{2}} d t \\
&=\frac{\sqrt{2}-2}{2} \int_{0}^{\infty} \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^{2}+(2+\sqrt{2})}+\frac{4-\sqrt{2}}{2} \int_{0}^{\infty} \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^{2}-(2 -\sqrt{2})}\\
&=\frac{\sqrt{2}-2}{2 \sqrt{\sqrt{2}+2}}\left[\tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{\sqrt{2}+2}}\right)\right]_{0}^{\infty}+0 \\
&=\frac{(\sqrt{2}-2) \pi}{2 \sqrt{\sqrt{2}+2}}
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
L &=-\frac{2+\sqrt{2}}{2} \int_{0}^{\infty} \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^{2}+(2-\sqrt{2})} =-\frac{(2+\sqrt{2}) \pi}{2 \sqrt{2-\sqrt{2}}}
\end{aligned}
$$
$$
\therefore J=\frac{1}{\sqrt{2}}\left[\frac{(\sqrt{2}-2) \pi}{2 \sqrt{\sqrt{2}+2}}+\frac{(2+\sqrt{2}) \pi}{2 \sqrt{2-\sqrt{2}}}\right] =\frac{\pi}{4} \sqrt{10-\sqrt{2}}$$
Hence we can conclude that
$$\boxed{I=\frac{\pi^{2}}{8} \sqrt{10-\sqrt{2}}}$$
How about when $n\geq 5$, $$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin ^{2n} x+\cos ^{2n} x} d x ?$$
Would you please help me?
Best Answer
Too long for a comment.
Very interested by the post, I played using the same approach and considered the more general case of
$$I_n=\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin ^{n} (x)+\cos ^{n} (x)} \,dx=\frac \pi 4\int_{0}^{\frac{\pi}{2}} \frac{dx}{\sin ^{n} (x)+\cos ^{n} (x)} $$ Using, as you did, $x=\tan ^{-1}(t)$, then $$I_n=\frac \pi 4\int_{0}^\infty \frac{\left(t^2+1\right)^{\frac{n}{2}-1}}{t^n+1}\,dt$$ The method does not work for odd values of $n$ (this is normal since the solution is given in terms of Meijer G-functions). But, for even values of $n$, we need one more integral each time and the result, is given by $$J_m=I_{2n}=\frac \pi 4\int_{0}^\infty \frac{\left(t^2+1\right)^{m-1}}{t^{2m}+1}\,dt=\frac {\pi^2} 4 a_m$$ The very first $a_m$ are $$\left\{\frac{1}{2},\frac{1}{\sqrt{2}},1,\frac{\sqrt{10-\sqrt{2}}}{2},\sqrt{5},\frac{1}{3} \sqrt{\frac{379}{2}-44 \sqrt{3}}\right\}$$ Now, using a CAS, what is interesting is that, when $m>6$, if $m$ is odd, the next ones are given by the solution of polynomial equations (cubic for $m=7,9$, quintic for $m=11$, sextic for $m=13$ and so on).
For example, to obtain $J_7$, we need to solve $y^3-10 y^2-32 y+328=0$ which makes $$J_7=\frac{1}{6} \left(5+14 \sin \left(\frac{1}{3} \sin ^{-1}\left(\frac{71}{98}\right)\right)\right)$$