Spurred on this question I decided to investigate the following integral:
\begin{equation}
I_n = \int_0^{\frac{\pi}{2}}\frac{1}{\sin^{2n}(x) + \cos^{2n}(x)}\:dx
\end{equation}
Where $n \in \mathbb{N}$.
The approach I've taken is rather simple and whilst I've arrived at a closed form solution, I'm wondering whether the resultant sum can be expressed in terms of (non)-elementary functions. Would love to see other methods that can be used to solve this! (Using any methods).
First:
\begin{equation}
I_n = \int_0^{\frac{\pi}{2}}\frac{1}{\sin^{2n}(x) + \cos^{2n}(x)}\:dx = \int_0^{\frac{\pi}{2}}\frac{1}{\left[\sin^{2}(x)\right]^n + \left[\cos^{2}(x)\right]^n}\:dx
\end{equation}
Using the Double-Angle Formulas:
\begin{equation}
\sin^2(x) = \frac{1 – \cos\left(2x\right)}{2} \qquad \cos^2(x) = \frac{1 + \cos\left(2x\right)}{2}
\end{equation}
Thus:
\begin{align}
I_n &= \int_0^{\frac{\pi}{2}}\frac{1}{\left[\sin^{2}(x)\right]^n + \left[\cos^{2}(x)\right]^n}\:dx = \int_0^{\frac{\pi}{2}}\frac{1}{\left[\frac{1 – \cos\left(2x\right)}{2}\right]^n + \left[\frac{1 + \cos\left(2x\right)}{2}\right]^n}\:dx \\
&= 2^n \int_0^{\frac{\pi}{2}} \frac{1}{\left[1 – \cos\left(2x\right)\right]^n + \left[1 + \cos\left(2x\right)\right]^n}\:dx
\end{align}
Making the substitution $u = 2x$
\begin{equation}
I_n = 2^n \int_0^{\frac{\pi}{2}} \frac{1}{\left[1 – \cos\left(2x\right)\right]^n + \left[1 + \cos\left(2x\right)\right]^n}\:dx = 2^{n – 1} \int_0^{\pi} \frac{1}{\left[1 – \cos\left(u\right)\right]^n + \left[1 + \cos\left(u\right)\right]^n}\:du
\end{equation}
Now apply the Weierstrass (/Tangent half angle) substitution $t = \tan\left(\frac{u}{2}\right)$:
\begin{align}
I_n &= 2^{n – 1} \int_0^{\pi} \frac{1}{\left[1 – \cos\left(u\right)\right]^n + \left[1 + \cos\left(u\right)\right]^n}\:du = 2^{n – 1} \int_0^{\infty} \frac{1}{\left[1 – \frac{1 – t^2}{1 + t^2}\right]^n + \left[1 + \frac{1 – t^2}{1 + t^2}\right]^n} \frac{2\:dt}{t^2 + 1} \\
&= \int_0^{\infty} \frac{\left[1 + t^2 \right]^{n – 1}}{t^{2n} + 1}\:dt
\end{align}
By the Binomial Theorem:
\begin{equation}
\left[1 + t^2 \right]^{2n – 1} = \sum_{j = 0}^{n – 1} {n – 1 \choose j}t^{2j}
\end{equation}
Thus:
\begin{align}
I_n &= \int_0^{\infty} \frac{\left[1 + t^2 \right]^{2n – 1}}{t^{2n} + 1}\:dt = \sum_{j = 0}^{n – 1} {n – 1 \choose j} \int_0^{\infty} \frac{t^{2j}}{t^{2n} + 1}
\end{align}
Using the solution I found here we find:
\begin{align}
I_n &= \sum_{j = 0}^{n – 1} {n – 1 \choose j} \int_0^{\infty} \frac{t^{2j}}{t^{2n} + 1} = \sum_{j = 0}^{n – 1} {n – 1 \choose j} \cdot \frac{1}{2n} \cdot 1^{\frac{2j + 1}{2n} – 1}B\left(1 – \frac{2j + 1}{2n} , \frac{2j + 1}{2n}\right) \\
&= \frac{1}{2n}\sum_{j = 0}^{n – 1} {n – 1 \choose j} B\left(1 – \frac{2j + 1}{2n} , \frac{2j + 1}{2n}\right)
\end{align}
Using the relationship between the Beta and Gamma function:
\begin{align}
I_n &= \frac{1}{2n}\sum_{j = 0}^{n – 1} {n – 1 \choose j} B\left(1 – \frac{2j + 1}{2n} , \frac{2j + 1}{2n}\right) = \frac{1}{2n}\sum_{j = 0}^{n – 1} {n – 1 \choose j} \frac{\Gamma\left(1 – \frac{2j + 1}{2n}\right)\Gamma\left(\frac{2j + 1}{2n}\right)}{\Gamma\left(1 – \frac{2j + 1}{2n} + \frac{2j + 1}{2n}\right)} \\
&= \frac{1}{2n}\sum_{j = 0}^{n – 1} {n – 1 \choose j}\Gamma\left(1 – \frac{2j + 1}{2n}\right)\Gamma\left(\frac{2j + 1}{2n}\right)
\end{align}
As $\frac{2j + 1}{2n} \not \in \mathbb{Z}$ we can employ Euler's Reflection Formula to yield:
\begin{align}
I_n &= \frac{1}{2n}\sum_{j = 0}^{n – 1} {n – 1 \choose j}\Gamma\left(1 – \frac{2j + 1}{2n}\right)\Gamma\left(\frac{2j + 1}{2n}\right) = \frac{1}{2n}\sum_{j = 0}^{n – 1} {n – 1 \choose j} \frac{\pi}{\sin\left(\pi \cdot \frac{2j + 1}{2n}\right)} \\&= \frac{\pi}{2n}\sum_{j = 0}^{n – 1} {n – 1 \choose j} \operatorname{cosec}\left(\frac{\pi}{2n}\left(2j + 1\right) \right)
\end{align}
Thus,
\begin{equation}
\int_0^{\frac{\pi}{2}}\frac{1}{\sin^{2n}(x) + \cos^{2n}(x)}\:dx = \frac{\pi}{2n}\sum_{j = 0}^{n – 1} {n – 1 \choose j} \operatorname{cosec}\left(\frac{\pi}{2n}\left(2j + 1\right) \right)
\end{equation}
Edit – Realised that $2n – 1$ should be $n – 1$
Extra Addition:
Using the same method as above an by letting
\begin{equation}
S_n(x) = \sin^{2n}(x) + \cos^{2n}(x)
\end{equation}
It becomes rather easy to solve:
\begin{equation}
I_{n,m} = \int_0^{\frac{\pi}{2}} \frac{S_n(x)}{S_m(x)}\:dx
\end{equation}
Where $n \lt m$. After applying the double angle formulas, $u$-substitution, and half tangent formula we arrive at:
\begin{equation}
I_{n,m} = \int_0^{\infty} \frac{S_n(x)}{S_m(x)}\:dx = \int_0^{\infty} \frac{t^{2n} + 1}{t^{2m} + 1} \left[1 + t^2 \right]^{m – n – 1}\:dt
\end{equation}
Applying the Binomial Expansion:
\begin{align}
I_{n,m} &= \sum_{j = 0}^{n – 1} {m – n – 1 \choose j} \int_0^{\infty} \frac{t^{2n} + 1}{t^{2m} + 1} \cdot t^{2j}\:dt \\
&= \sum_{j = 0}^{n – 1} {m – n – 1 \choose j} \int_0^{\infty} \frac{t^{2\left(n + j\right)} }{t^{2m} + 1} \:dt + \sum_{j = 0}^{n – 1} {m – n – 1 \choose j} \int_0^{\infty} \frac{t^{2j} }{t^{2m} + 1} \:dt\\
\end{align}
Again applying another of my solutions (referenced above) we arrive at:
\begin{align}
I_{n,m} &= \sum_{j = 0}^{n – 1} {m – n – 1 \choose j} \frac{\pi}{2m}\operatorname{cosec}\left(\frac{\pi}{2m}\left(2\left(n + j\right) + 1\right) \right) + \sum_{j = 0}^{n – 1} {m – n – 1 \choose j}\frac{\pi}{2m}\operatorname{cosec}\left(\frac{\pi}{2m}\left(2j + 1\right) \right) \\
&= \frac{\pi}{2m}\sum_{j = 0}^{n – 1} {m – n – 1 \choose j}\left[\operatorname{cosec}\left(\frac{\pi}{2m}\left(2\left(n + j\right) + 1\right) \right) + \operatorname{cosec}\left(\frac{\pi}{2m}\left(2j + 1\right) \right) \right]
\end{align}
Note: Using the same method we can easily solve
\begin{equation}
I_{n,p} = \int_0^{\frac{\pi}{2}}\frac{1}{\left(\sin^{2n}(x) + \cos^{2n}(x) \right)^p}\:dx = \frac{1}{2n\Gamma(p)}\sum_{m = 0}^{np – 1} {np – 1 \choose m}\Gamma\left(p – \frac{2m + 1}{2n}\right)\Gamma\left(\frac{2m + 1}{2n}\right)
\end{equation}
Where $p \in \mathbb{N}$
Best Answer
I cannot seem to add much here other than a slightly quicker way for you to reach the integral you arrive at after making a tangent half-angle substitution.
Factoring out a $\cos^{2n} x$ term in the denominator of the original integral for $I_n$ we have $$I_n = \int_0^{\frac{\pi}{2}} \frac{\sec^{2n} x}{1 + \tan^{2n} x} \, dx = \int_0^{\frac{\pi}{2}} \frac{\sec^{2n - 2} x}{1 + \tan^{2n} x} \sec^2 x \, dx.$$ On setting $t = \tan x, dt = \sec^2 x \, dx$, we have $$I_n = \int_0^\infty \frac{(1 + t^2)^{n - 1}}{1 + t^{2n}} \, dt.$$