# Can we evaluate $\int \frac{1}{\sin ^ {2n+1} x+\cos ^ {2n+1} x} d x?$

After investigating the integral

$$\boxed{\quad \int \frac{1}{\sin ^5 x+\cos ^5x} d x \\=\frac{4}{5}\left[-\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{2}}\right)+\frac{1}{\sqrt{\sqrt 5-2}} \tan ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}-2}}\right)\\ \qquad\ -\frac{1}{\sqrt{\sqrt{5}+2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}+2}}\right)\right]+C}$$

in my post, I try to evaluate $$\int \frac{1}{\sin ^ {7} x+\cos ^ {7} x} d x$$ using the factorisation

\begin{aligned} \sin ^{7} x+\cos ^7 x&=\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^5 x+\cos ^5 x\right)-\sin ^2 x \cos ^2 x\left(\sin ^3 x+\cos ^3 x\right) \\ &=\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^3 x+\cos ^3 x\right) -\sin ^2 x \cos ^2 x(\sin x+\cos x) \\ &\quad -\sin ^2 x \cos ^2 x\left(\sin ^3 x+\cos ^3 x\right) \\ &=(\sin x+\cos x)\left(1-\sin x \cos x-2 \sin ^2 x \cos ^2 x+\sin^3x\cos^3x\right) \end{aligned}

Let $$t=\sin x-\cos x$$, then $$d t=(\cos x+\sin x) d x$$ and $$\sin x\cos x= \frac{1-t^2}{2}$$ and yields \begin{aligned} I =\int \frac{8}{\left(t^2-2\right)\left(t^6+t^4-9 t^2-1\right)} d t \end{aligned} where I was stuck in the last integrand with power $$6$$. Your help and comments are highly appreciated.

My Question:

Is it difficult go further with $$\int \frac{1}{\sin ^ {2n+1} x+\cos ^ {2n+1} x} d x?$$ where $$n\geq 3.$$

• How is it?$$\sin ^7 x+\cos ^7 x =\left(\sin ^3 x+\cos ^3 x\right)\left(\sin ^4 x+\cos ^4 x\right)$$ Commented Feb 3, 2023 at 15:23
• You are right, I am trying to fix it!
– Lai
Commented Feb 4, 2023 at 0:06

## 1 Answer

With $$a_k=\frac{\pi k}{2n+1}$$, decompose the integrand as follows \begin{align} &\frac{2n+1}{\sin^{2n+1}x+\cos^{2n+1}x}\\ =& \ \frac{2^n}{\sin x+\cos x}+2^{n+1}\sum_{k=1}^n(-1)^{k}\frac{\cos^{n}2a_k}{\sec a_k }\frac{\sin x+\cos x}{1+\cos 2a_k\sin2x} \\ \end{align} Then, the RHS can be integrated piece-wise. For example

\begin{align} &\frac78\int \frac1{\sin^{7} x+\cos^{7} x}dx\\ =&\int \frac{1}{2-t^2} - \frac{2\cos\frac\pi7 \cos^2\frac{2\pi}7}{{\sec\frac{2\pi}7+1-t^2}} +\frac{2\cos\frac{2\pi}7 \cos^2\frac{4\pi}7}{{\sec\frac{4\pi}7+1-t^2}} -\frac{2\cos\frac{3\pi}7 \cos^2\frac{6\pi}7}{{\sec\frac{6\pi}7+1-t^2}}\ dt \end{align} where the substitution $$t=\sin x-\cos x$$ is made.