How to solve this definite integral $\int_0^\pi \frac{\cos^9(x)}{\sin^3(x)+\cos^3(x)}dx$ I'm having trouble evaluating the following integral:
$$
\int^\pi_0 \frac{\cos^9(x)}{\sin^3(x)+\cos^3(x)}dx
$$
I tried to convert it into an algebraic function by multiplying the numerator and denominator by $\sec^{11}(x)$ and substituting $\tan(x)=t$ as
$$
\int^\pi_0 \frac{\cos^9(x)\cdot \sec^{11}(x)}{(\sin^3(x)+\cos^3(x))\cdot \sec^{11}(x)}dx
$$
$$
\int^\pi_0 \frac{\sec^2(x)}{(\tan^3(x)+1)\cdot (\tan^2(x)+1)^4}dx
$$
Substituting $\tan(x)=t$,
$$
\int^0_0 \frac{dt}{(t^3+1)\cdot (t^2+1)^4}
$$
But now both upper and lower limit become $0$ so apparently this is not the right approach, so how do I go about solving it?
 A: Assuming the usage of the Cauchy principle value, let's define
$$ I_1 := \int_0^\frac\pi2 \frac{\cos^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
\\ I_2 := \text{p. v.} \int_\frac\pi2^\pi \frac{\cos^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
$$
For $I_1$, by making a substitution and adding it to itself,
$$
I_1 = \int_0^\frac\pi2 \frac{\sin^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
\\ = \frac12 \int_0^\frac\pi2 \frac{\cos^9(x) +\sin^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
\\ = \frac12 \int_0^\frac\pi2 \left( \sin^6(x) + \cos^6(x) - \sin^3(x)\cos^3(x) \right) {\rm d}x
$$
For $I_2$, by making some substitutions,
$$
I_2 = \text{p. v.} \int_0^\frac\pi2 \frac{\sin^9(x)}{\sin^3(x)-\cos^3(x)} {\rm d}x
\\  = \text{p. v.} \int_0^\frac\pi2 \frac{-\cos^9(x)}{\sin^3(x)-\cos^3(x)} {\rm d}x
\\  = \frac12 \text{p. v.} \int_0^\frac\pi2 \frac{\sin^9(x)-\cos^9(x)}{\sin^3(x)-\cos^3(x)} 
\\  = \frac12 \int_0^\frac\pi2 \left( \sin^6(x) + \cos^6(x) + \sin^3(x)\cos^3(x) \right) {\rm d}x
$$
Therefore,
$$
I_1 + I_2 = \text{p. v.} \int_0^\pi \frac{\cos^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
\\ = \int_0^\frac\pi2 \sin^6(x) + \cos^6(x) {\rm d}x
\\ = \int_0^\frac\pi2 \frac58 + \frac38 \cos(4x) {\rm d}x
\\ = \frac{5\pi}{16}
$$
