Given $$\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx$$
First we will simplify $$\sin^5 x+\cos^5 x = \left(\sin^2 x+\cos^2 x\right)\cdot \left(\sin^3 x+\cos^3 x\right) - \sin ^2x\cdot \cos ^2x\left(\sin x+\cos x\right)$$
$$\displaystyle \sin^5 x+\cos^5 x= (\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)$$
So Integral is $$\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx $$
$$\displaystyle = \int\frac{1}{(\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$
$$\displaystyle = \int \frac{(\sin x+\cos x)}{(\sin x+\cos x)^2\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$
$$\displaystyle = \int \frac{(\sin x+\cos x)}{(1+\sin 2x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$
Let $$(\sin x-\cos x) = t\;,$$ Then $$(\cos +\sin x)dx = dt$$ and $$(1-\sin 2x) = t^2\Rightarrow (1+\sin 2x) = (2-t^2)$$
So Integral Convert into $$\displaystyle = 4\int\frac{1}{(2-t^2)\cdot(5-t^4)}dt = 4\int\frac{1}{(t^2-2)\cdot (t^2-\sqrt{5})\cdot (t^2+\sqrt{5})}dt$$
Now Using partial fraction, we get
$$\displaystyle = 4\int \left[\frac{1}{2-t^2}+\frac{1}{(2-\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}-t^2)}+\frac{1}{(2+\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}+t^2)}\right]dt$$
$$ = \displaystyle \sqrt{2}\ln \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\frac{1}{(2-\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \ln \left|\frac{5^{\frac{1}{4}}+t}{5^{\frac{1}{4}}-t}\right|+\frac{2}{(2+\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \tan^{-1}\left(\frac{t}{5^{\frac{1}{4}}}\right)+\mathbb{C}$$
where $$t=(\sin x-\cos x)$$
Let
\begin{equation*}
I=\int \cos 2x\cdot\ln \left|\frac{\cos x+\sin x}{\cos x-\sin x}\right|\,dx.
\end{equation*}
Using the following identity
\begin{equation*}
\cos 2x=2\cos ^{2}x-1
\end{equation*}
and the substitution
\begin{equation*}
u=\cos x,
\end{equation*}
we get
\begin{equation*}
I=\int \frac{1-2u^{2}}{\sqrt{1-u^{2}}}\cdot\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|\,du.
\end{equation*}
$I$ is integrable by parts, differentiating the factor $\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|$ and integrating the factor $\frac{1-2u^{2}}{\sqrt{1-u^{2}}}$. After simplifying, we obtain
\begin{eqnarray*}
I &=&u\sqrt{1-u^{2}}\cdot\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|+2\int
\frac{u}{2u^{2}-1}du \\[2ex]
&=&u\sqrt{1-u^{2}}\cdot\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|+\frac{1}{2}
\ln \left| 2u^{2}-1\right| +C \\[2ex]
&=&\left( \cos x\cdot\sin x\right)\cdot \ln \left|\frac{\cos x+\sin x}{\cos x-\sin x}\right|+\frac{1
}{2}\ln \left| 2\cos ^{2}x-1\right| +C\\[2ex]
&=&\frac{\sin 2x
}{2} \ln \left|\frac{\cos x+\sin x}{\cos x-\sin x}\right|+\frac{\ln \left| \cos 2x\right|
}{2} +C.
\end{eqnarray*}
Best Answer
Hint:
I found the expression $$=\dfrac{\sin(x-1)}{\cos(x-1)\cos(x-3)}-\dfrac{\sin(x-2)}{\cos(x-2)\cos(x-3)}$$
Now,
$$\dfrac{\sin(x-1)}{\cos(x-1)\cos(x-3)}$$
$$=\dfrac{\sin(x-1)}{\sin2}\cdot\dfrac{\sin(x-1-(x-3))}{\cos(x-1)\cos(x-3)}$$
$$=\dfrac{1-\cos^2(x-1)}{\sin2\cos(x-1)}-\dfrac{\sin(x-1)\sin(x-3)}{\sin2\cos(x-3)}$$
The first part can be managed easily.
For the second part, $$\dfrac{\sin(x-1)\sin(x-3)}{\cos(x-3)}=\dfrac{\sin(x-3+2)\sin(x-3)}{\cos(x-3)}=\dfrac{\cos2\sin^2(x-3)}{\cos(x-3)}-\sin2\sin(x-3)$$
Can you take it home from here?