Is it easy to evaluate the integral $\int \frac{\sin ^{5} x+\cos ^{5} x}{\sin ^{3} x+\cos ^{3} x} d x$

Solution 1:

Let $$ I_{n}:=\int \frac{\sin ^{n} x+\cos ^{n} x}{\sin ^{3} x+\cos ^{3} x} d x $$ After getting the answer for the integral $I_5$, I started to investigate the integral with higher powers $n$. In a couple of days, I found a reduction formula for it.

For any natural number $n\geq 4$, we have $$ \begin{aligned} \sin ^{n} x+\cos ^{n} x=&\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{n-2} x+\cos ^{n-2} x\right) -\sin ^{2} x \cos ^{n-2} x-\cos ^{2} x \sin ^{n-2} x. \end{aligned} $$ $$ \frac{\sin ^{n} x+\cos ^{n} x}{\sin ^{3} x+\cos ^{3} x}=\frac{\sin ^{n-2} x+\cos ^{n-2} x}{\sin ^{3} x+\cos ^{3} x}-\frac{\sin ^{2} x \cos ^{2} x\left(\sin ^{n-4} x+\cos ^{n-4} x\right)}{\sin ^{3} x+\cos ^{3} x} $$ We got a reduction formula for $I_n.$ $$ I_{n}=I_{n-2}-\int \frac{\sin ^{2} x \cos ^{2} x\left(\sin ^{n-4} x+\cos ^{n-4} x\right)}{\sin ^{3} x+\cos ^{3} x} d x $$

As an example, let’s start with $I_7$ $$ \begin{aligned}I_{7}&=I_{5}-\int \sin ^{2} x \cos ^{2} x d x\\&= 2x+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1-2 \tan x}{\sqrt{3}}\right)-\frac{\cos (2 x)}{4} -\frac{1}{4} \int \frac{1-\cos (4 x)}{2} d x\\&= \frac{15x}{8} +\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1-2 \tan x}{\sqrt{3}}\right)-\frac{\cos (2 x)}{4} +\frac{\sin (4x)}{32} +C \end{aligned}$$ For $I_9,$ we use division instead of the reduction formula.

$$ \begin{aligned} I_{9} &=\int \frac{\left(\sin ^{3} x+\cos ^{3} x\right)\left(\sin ^{6} x-\sin ^{3} x \cos ^{3} x+\cos ^{6} x\right)}{\sin ^{3} x+\cos ^{3} x} d x \\ &=\underbrace{\int \left(\sin ^{6} x+\cos ^{6} x \right) d x}_{J}-\underbrace{\int \sin ^{3} x \cos ^{3} x d x}_{K} \end{aligned} $$

For the integral $J$, we have

$$ \begin{aligned} \sin ^{6} x+\cos ^{6} x &=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right) \\ &=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-3 \sin ^{2} x \cos ^{2} x \\ &=1-\frac{3}{4} \sin ^{2}(2 x) \\ &=1-\frac{3}{8}(1-\cos (4 x) )\\ &=\frac{1}{8}(5+3 \cos (4 x)) \end{aligned} $$ We get $$ J=\frac{1}{8}\left(5 x+\frac{3 \sin (4 x)}{4}\right)+C_{1} $$

For $K$, using the double-angle formula yields

$$ \begin{aligned} K &=\int \sin ^{3} x \cos ^{3} x d x \\ &=\frac{1}{8} \int \sin ^{3}(2 x) d x \\ &=-\frac{1}{16} \int\left(1-\cos ^{2}(2 x)\right) d(\cos 2 x) \\ &=-\frac{1}{16}\left(\cos (2 x)-\frac{\cos ^{3}(2 x)}{3}\right)+C_2 \end{aligned} $$

Now we can conclude that

$$ I_{9}=\frac{1}{96}\left[60 x+9 \sin (4 x)-6 \cos (2 x)+2 \cos ^{3}(2 x)\right]+C $$

By the reduction formula, we now dare to find $I_{11}.$

$$ \begin{aligned} I_{11} &=I_{9}-\int \frac{\sin ^{2} x \cos ^{2} x\left(\sin ^{5} x+\cos ^{5} x\right)}{\sin ^{3} x+\cos ^{3} x} d x \\ &=I_{9}-\int \sin ^{2} x \cos ^{2} x\left(1-\frac{\sin ^{2} x \cos ^{2} x}{1-\sin x \cos x}\right) d x \\ &=I_{9}- \underbrace{ \int \sin ^{2} x \cos ^{2} x d x}_{L}+\underbrace{\int \frac{\sin ^{4} x \cos ^{4} x}{1-\sin x \cos x}}_{M} d x \end{aligned} $$

$$ \begin{aligned} L &=\frac{1}{4} \int \sin ^{2}(2 x) d x=\frac{1}{8} \int(1-\cos 4 x) d x =\frac{1}{8}\left(x-\frac{\sin (4 x)}{4}\right)+C_{1} \end{aligned} $$

Using the formula in my answer, we have

$$ \begin{aligned} M=& \int \frac{\sin ^{4} x \cos ^{4} x}{1-\sin x \cos x} d x \\ =& \int \frac{\sin ^{3} x \cos ^{3} x}{1-\sin x \cos x} d x-\int \sin ^{3} x \cos ^{3} x d x \\ =& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-\frac{9}{8} x+\frac{\cos 2 x}{4}+\frac{\sin 4 x}{32} +\frac{1}{16}\left[\cos (2 x)-\frac{\cos ^{3}(2 x)}{3}\right]+C . \\ =& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-\frac{9}{8} x +\frac{\sin 4 x}{32} +\frac{5}{16} \cos (2 x)-\frac{1}{48} \cos ^{3}(2 x)+C \end{aligned} $$

Grouping the integral together yields

$$ \begin{aligned} I_{11}=& \frac{1}{96}\left[-36 x+15 \sin (4 x)+29\cos (2 x)\right] +\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C \end{aligned} $$

Can you help me go further?