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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?

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    $\begingroup$ Before you do anything, you need to decide what you want to do with the singularity at $x = \frac{3\pi}{4}$. $\endgroup$ Commented Jan 16, 2016 at 19:20
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    $\begingroup$ One option is compute the Cauchy principal value instead. Another option is lift the integral to a contour integral in complex plane and deform the contour around the pole at $x = \frac{3\pi}{4}$. If this is the first time you deal with this sort of integral, you should follow the Cauchy principle value route. You essentially need to split the integral to 2 regions $[0,\frac{3\pi}{4}-\epsilon], [\frac{3\pi}{4}+\epsilon,\pi]$ and take the limit $\epsilon \to 0^{+}$ at the end. $\endgroup$ Commented Jan 16, 2016 at 19:32
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    $\begingroup$ Sure the integral is not on $[0,\pi/2]$? $\endgroup$
    – Did
    Commented Jan 16, 2016 at 20:19
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    $\begingroup$ "no harm in learning something new" Sure, and this is related how? $\endgroup$
    – Did
    Commented Jan 17, 2016 at 10:34
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    $\begingroup$ When you do your substitutions, the functions in your substitutions must be continuous over the required interval. In this case, break it up into $(0, \frac {\pi}{2}), (\frac {\pi}{2}, \pi)$ and you should be fine. $\endgroup$
    – Doug M
    Commented Dec 20, 2018 at 18:42

1 Answer 1

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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,

$$ I_1 + I_2 = \text{p. v.} \int_0^\pi \frac{\cos^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x \\ = \int_0^\frac\pi2 \sin^6(x) + \cos^6(x) {\rm d}x \\ = \int_0^\frac\pi2 \frac58 + \frac38 \cos(4x) {\rm d}x \\ = \frac{5\pi}{16} $$

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