MHB What is the Integral of a Trigonometric Expression with Positive Constants?

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Here is this week's POTW:

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Find, with proof, the exact value of

$$\int_0^{\pi/2} \frac{\cos^3u\sin^7 u}{(p\cos^2u + q\sin^2u)^{6}}\, du$$

where $p$ and $q$ are positive constants.-----

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No one answered this week's problem. Here is my solution.

Spoiler

The integral evaluates to $\frac{1}{40p^2q^4}$. Let $I(p,q)$ denote the value of the integral. By a change of variable $x = \sin^2 u$,

$$I(p,q) = \frac{1}{2}\int_0^1 \frac{(1 - x)x^3}{[(q - p)x + p]^6}\, dx.$$

If $p = q$, then

$$I(p,q) = \frac{1}{2}\int_0^1 \frac{x^3 - x^4}{p^6}\, dx = \frac{1}{40p^6}.$$

If $p\neq q$, then

$$I(p,q) = \frac{1}{2(q-p)^6} \int_0^1 \frac{(1 - x)x^3}{(x + r)^6}\, dx$$

where $r = \frac{p}{q - p}$. Now let $y = \frac{(r+1)x}{x+r}$.Then

$$I(p,q) = \frac{1}{2(q-p)^6}\int_0^1 \left(\frac{1 - x}{x + r}\right)^2\left(\frac{x}{x + r}\right)^4 \frac{1}{x(1 - x)} dx = \frac{1}{2(q - p)^6}\int_0^1 \left(\frac{1-y}{r}\right)^2\left(\frac{y}{r+1}\right)^4\frac{(r + 1 - y)^2}{r(r+1)y(1-y)}\frac{r(r+1)}{(r + 1 - y)^2}\, dy$$
$$\qquad = \frac{1}{2(q-p)^6r^2(r + 1)^4}\int_0^1 (1 - y)y^3\, dy$$
$$\qquad = \frac{1}{40p^2q^4}.$$