Seeking help to find the exact value of $ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin ^{2n} x+\cos ^{2n} x} d x $ using substitutions?

Question

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

$$ \boxed{I_{n}=\frac{\pi^2}{8 n} \sum_{k=0}^{n-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) \csc \frac{(2 k+1) \pi}{2 n}} $$

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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: $$ \boxed{I_{n}=\frac{\pi^2}{8 n} \sum_{k=0}^{n-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) \csc \frac{(2 k+1) \pi}{2 n}} $$

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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?

Answer 1

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 $$J_7=\frac{1}{6} \left(5+14 \sin \left(\frac{1}{3} \sin ^{-1}\left(\frac{71}{98}\right)\right)\right)$$

Answer 2

This is where I got with complex integration.

$\frac {\pi}{2}\int_0^{\infty} \frac {(t^2 + 1)^3}{t^8 + 1} = \frac {\pi}{4}\int_{-\infty}^{\infty} \frac {(t^2 + 1)^3}{t^8 + 1}$

Using the contour of the the semicircle in the the upper half plane....

$\lim_\limits{R\to \infty} \left|\frac {(R^2 + 1)^3}{R^8 + 1}R\right| = 0$

The integral along the circular part of the path equals 0.

There are 4 poles inside the contour. Evaluating the residues....

$\frac {\pi}{4}(2\pi i) \left(\frac{(e^{i\frac {\pi}{4}} + 1)^3}{8e^{\frac{7\pi}{8}i}} +\frac{(e^{i\frac {3\pi}{4}} + 1)^3}{8e^{\frac {21\pi}{8}i}}+\frac{(e^{i\frac {5\pi}{4}} + 1)^3}{8e^{\frac {35\pi}{8}i}} + \frac{(e^{i\frac {7\pi}{4}} + 1)^3}{8e^{\frac {49\pi}{8}i}}\right)$

$\frac {\pi}{32}(2\pi i) \left(-e^{\frac {\pi}{8}i}(e^{\frac {\pi}{4}i} + 1)^3 -e^{\frac {3\pi}{8}i}(e^{\frac {3\pi}{4}i} + 1)^3 - e^{\frac {5\pi}{8}i}(e^{\frac {5\pi}{4}i} + 1)^3 - e^{\frac {7\pi}{8}i}(e^{\frac {7\pi}{4}i} + 1)^3\right)$

$e^{ki}(e^{2ki} + 1)^3 = e^{ki}(e^{6ki} + 3e^{4ki} + 3e^{2ki} + 1) = e^{7ki} + 3e^{5ki} + 3e^{3ki} + e^{ki}$

$(-\frac {\pi^2}{16}i)(2e^{\frac{\pi}{8}i} + 8e^{\frac{3\pi}{8}i} + 8e^{\frac{5\pi}{8}i} + 2e^{\frac{7\pi}{8}i}+6e^{\frac{9\pi}{8}i} + 6e^{\frac{15\pi}{8}i})$

$e^{ix} + e^{(\pi - x)i} = 2i\sin x\\ 2e^{\frac{\pi}{8}i} + 2e^{\frac{7\pi}{8}i} + 8e^{\frac{3\pi}{8}i} + 6e^{\frac{9\pi}{8}i} + 6e^{\frac{15\pi}{8}i} = 4i\sin\frac {\pi}{8} + 16i\sin \frac {3\pi}{8} - 12 i \sin \frac{\pi}{8}$

$(\frac {\pi^2}{16})(16\sin \frac {3\pi}{8} - 8 \sin \frac{\pi}{8})$

$(\frac {\pi^2}{16})(8\sqrt {2+\sqrt 2} - 4\sqrt {2-\sqrt 2})$
$(\frac {\pi^2}{4})(2\sqrt {2+\sqrt 2} - \sqrt {2-\sqrt 2})$

$\sqrt {10 - \sqrt 2} = (2\sqrt {2+\sqrt 2} - \sqrt {2-\sqrt 2})$

Per David's suggestion,

Suppose we take the contour from $0$ to $R$ along the real line, from $R$ to $Re^{\frac {\pi}{4}i}$ along the arc, and back to $0$ in a line.

$\int_0^R\frac {(t^2 + 1)^3}{t^8+1}\ dt = \int_0^R\frac {t^6}{t^8+1}\ dt + 3\int_0^R\frac {t^4}{t^8+1}\ dt+ 3\int_0^R\frac {t^2}{t^8+1}\ dt + \int_0^R\frac {1}{t^8+1}\ dt$

$\int_0^R \frac {t^n}{t^8 + 1}\ dt + \int_0^R \frac {(e^{\frac {\pi}{4}i} t)^n}{(e^{\frac {\pi}{4}i} t)^8 + 1}\ d(e^{\frac {\pi}{4}i} t)\\ (1-e^{\frac {(n+1)\pi}{4}i})\int_0^R \frac {t^n}{t^8 + 1}\ dt$

for $n\le 6$
$\lim_\limits{R\to \infty} \int_0^R \frac {t^n}{t^8 + 1}\ dt = \frac {2\pi i}{1-e^{\frac {(n+1)\pi}{4} i}} \text { Res}_{z=e^{\frac {\pi}{8}i}} \left(\frac {z^n}{z^8 + 1}\right)$

$\frac {2\pi i}{1-e^{\frac {(n+1)\pi}{4} i}} \left(\frac {e^{\frac {n\pi}{8}i}}{8e^{\frac {7\pi}{8}i}}\right)$

$\frac {2\pi i}{8\left(e^{\frac {(7-n)\pi}{8}i}-e^{\frac {(n+9)\pi}{8} i}\right)}$
$\frac {2\pi i}{8\left(2i\sin \frac {(7-n)\pi}{8}\right)}$
$\lim_\limits{R\to \infty} \int_0^R \frac {t^n}{t^8 + 1}\ dt = \frac {\pi}{8}\csc \frac {(7-n)\pi}{8}$

$\frac {\pi}{2}\int_0^{\infty} \frac {(t^2 + 1)^3}{t^8 + 1} = \frac {\pi^2}{16}(\csc \frac {\pi}{8} + 3 \csc \frac {3\pi}{8} + 3 \csc \frac {5\pi}{8} + \csc \frac {7\pi}{8}) $

$\frac {\pi^2}{4}\sqrt{10-\sqrt 2}$

Answer 3

Utilize the known integral $\int_0^\infty \frac{t^{n-1}}{1+t^m}=\frac\pi m\csc\frac{\pi n}m$ \begin{align} I=&\>\frac{\pi}{4} \int_{0}^{\infty} \frac{\left(1+t^{2}\right)^{3}}{t^{8}+1} d t\\ =&\>\frac{\pi^2}{32}\left( \csc\frac\pi8+3\csc\frac{3\pi}8 +3\csc\frac{5\pi}8+\csc\frac{7\pi}8 \right)\\ =&\>\frac{\pi^2}{16}\left( \csc\frac\pi8+3\csc\frac{3\pi}8 \right) = \frac{\pi^2}8\sqrt{10-\sqrt2} \end{align}

Answer 4

An alternative approach:

We can use power-reducing formulas to get

\begin{align*} \sin^8(x)&=\frac1{128}(35-56\cos(2x)+28\cos(4x)-8\cos(6x)+\cos(8x))\\ \cos^8(x)&=\frac1{128}(35+56\cos(2x)+28\cos(4x)+8\cos(6x)+\cos(8 x)) \end{align*}

Therefore,

\begin{align*} \frac\pi4\int_0^{\frac\pi2}\frac{1}{\sin^8(x)+\cos^8(x)}\mathop{dx}&=16\pi\int_0^{\frac\pi2}\frac{1}{35+28\cos(4x)+\cos(8x)}\mathop{dx}\\ &=4\pi\int_0^{2\pi}\frac{1}{35+28\cos(x)+\cos(2x)}\mathop{dx}&(4x\mapsto x)\\ &=2\pi\int_0^{2\pi}\frac{1}{17+14\cos(x)+\cos^2(x)}\mathop{dx}\\ \end{align*}

Now, set $u=\tan\left(\frac x2\right)\implies du=\frac12\sec^2\left(\frac x2\right)\mathop{dx}$. From this, we can derive that $\cos(x)=\frac{1-u^2}{1+u^2}$ and $dx=\frac2{1+u^2}\mathop{du}$. So, after some simplifying, our integral becomes

\begin{align*} 2\pi\int_0^\infty\frac{1+u^2}{8+8u^2+u^4}\mathop{du}&=2\pi\int_0^\infty\frac{1+u^2}{(u^2+4+2\sqrt2)(u^2+4-2\sqrt2)}\mathop{du}\\ &=2\pi\int_0^\infty\frac{3+2\sqrt2}{4\sqrt2}\cdot\frac{1}{u^2+4+2\sqrt2}-\frac{3-2\sqrt2}{4\sqrt2}\cdot\frac{1}{u^2+4-2\sqrt2}\mathop{du}\\ &=\frac{\pi}{2\sqrt2}\left.\left[\frac{3+2\sqrt2}{\sqrt{4+2\sqrt2}}\arctan\left(\frac u{\sqrt{4+2\sqrt2}}\right)-\frac{3-2\sqrt2}{\sqrt{4-2\sqrt2}}\arctan\left(\frac u{\sqrt{4-2\sqrt2}}\right)\right]\right|_0^\infty\\ &=\frac{\pi^2}{4\sqrt2}\left[\frac{3+2\sqrt2}{\sqrt{4+2\sqrt2}}-\frac{3-2\sqrt2}{\sqrt{4-2\sqrt2}}\right]\\ &=\frac{\pi^2}{8}\sqrt{10-\sqrt2}\\ \end{align*}

Answer 5

The closed form for $$I_n= \frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{d x}{\sin ^{2n} x+\cos ^{2n} x} $$ is

$$ \boxed{I_{n}=\frac{\pi^2}{8 n} \sum_{k=0}^{n-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) \csc \frac{(2 k+1) \pi}{2 n}} $$

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Proof:

Letting $t=\tan x$ yields $$ \begin{aligned} I_{n} &=\frac{\pi}{4} \int_{0}^{\infty} \frac{\left(1+t^{2}\right)^{n-1}}{t^{2 n}+1} \\ &= \frac{\pi}{4} \sum_{k=0}^{n-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) \int_{0}^{\infty} \frac{t^{2 k}}{t^{2 n}+1} d t . \end{aligned} $$

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: $$ \boxed{I_{n}=\frac{\pi^2}{8 n} \sum_{k=0}^{n-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) \csc \frac{(2 k+1) \pi}{2 n}} $$

For examples:

\begin{aligned} I_{2}&=\frac{\pi^2}{16}\left(\csc \frac{\pi}{4}+\csc \frac{3 \pi}{4}\right)=\frac{\sqrt{2} \pi^{2}}{8} \\ I_{3}&=\frac{\pi^2}{24}\left(\csc \frac{\pi}{6}+2 \csc \frac{3 \pi}{6}+\csc \frac{5 \pi}{6}\right)=\frac{\pi^{2}}{4} \\ I_{4}&=\frac{\pi^2}{32} \left[\left(\begin{array}{l} 3 \\ 0 \end{array}\right) \csc \frac{\pi}{8}+\left(\begin{array}{l} 3 \\ 1 \end{array}\right) \csc \frac{3 \pi}{8}+\left(\begin{array}{l} 3 \\ 2 \end{array}\right) \csc \frac{5 \pi}{8}+\left(\begin{array}{l} 3 \\ 3 \end{array}\right) \csc \frac{6\pi}{8}\right]= \frac{\pi^2}8\sqrt{10-\sqrt2} \end{aligned}