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This integral:

$$\int_{0}^{\pi/2} \frac{y \sqrt{\cos(y)}}{\sqrt{\cos(y)} + \sqrt{\sin(y)}} dy$$

came up when I was playing around with trying to solve this similar problem:

Is there a closed form for $\int_0^{\pi/2} \dfrac{e^{-x}\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx $?

for which I had only a partial amount of success by Taylor-expanding $e^{-x}$ about $\frac{\pi}{4}$ - integrals of more generally

$$\int_{0}^{\pi/2} \frac{y^{2k+1} \sqrt{\cos(y)}}{\sqrt{\cos(y)} + \sqrt{\sin(y)}} dy$$

, $k = 0, 1, 2, \cdots$ are involved, so it'd be also interesting to hear about the general type too, but I'm wondering if cracking the simplest one wouldn't yield some hints at least.

The_Sympathizer
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  • I tried substituting $y' = \frac{\pi}{2} - y$ but it just gave me an obvious identity. Try integrating using parts, $y$ and the rest. I know a closed form of $f(y) = \int \frac{\sqrt{\cos y}}{\sqrt{\sin y}+\sqrt{\cos y}} dy$ - I'll leave that if you want. (But still it seems very difficult since you have to integrate that $f(y)$ again, and it is dirty work.) – Vue Apr 12 '22 at 11:58
  • Here's a link to $f(y)$ : https://math.stackexchange.com/questions/3535688/integrate-int-frac-sqrt-sin-x-sqrt-sin-x-sqrt-cos-xdx – Vue Apr 12 '22 at 12:04

1 Answers1

1

Disclaimer : this is not a full answer.

Let

$$I_{n} = \int_{0}^{\frac{\pi}{2}} \frac{y^{n} \sqrt{\cos y}}{\sqrt{\cos y} + \sqrt{\sin y}} dy$$

and

$$J_{n} = \int_{0}^{\frac{\pi}{2}} \frac{y^{n} \sqrt{\sin y}}{\sqrt{\cos y} + \sqrt{\sin y}} dy.$$

Substitute $y' = \frac{\pi}{2} - y$ then

$$I_{n} = \int_{0}^{\frac{\pi}{2}} \frac{(\pi/2 - y)^{n} \sqrt{\sin x}}{\sqrt{\cos y} + \sqrt{\sin x}} dy = (-1)^{n} \int_{0}^{\frac{\pi}{2}} \left( \sum_{i=0}^{n} \binom{n}{i} \left( - \frac{\pi}{2} \right)^{n-i} y^{i} \right) \frac{\sqrt{\sin y}}{\sqrt{\cos y} + \sqrt{\sin y}} dy$$

If $2 \:|\: n$ then we can take the $J_{n}$ of the RHS to the LHS, and we can evaluate $I_{n} - J_{n}$ using $J_{i} \:\:(i < n)$.

Meanwhile

$$I_{n} + J_{n} = \int_{0}^{\frac{\pi}{2}} y^{n} dy = \frac{\pi^{n+1}}{(n+1) \cdot 2^{n+1}}$$

so now we can evaluate $I_{n}$ using some recurrence relations of $J_{n}$. Unfortunately, this method is useless for $2 \not|\:n$, because when we take the $J_{n}$ to the LHS then it just becomes $I_{n} + J_{n}$. (It's the same reason why it doesn't work for $I_{1}$, it becomes an obvious identity and there's nothing we can do with it.)

Or maybe we can integrate using parts, where

$$\int \frac{y^{2k+1} \sqrt{\cos y}}{\sqrt{\cos y} + \sqrt{\sin y}} \cdot 1 dy = \frac{1}{2k+2} \cdot \frac{y^{2k+2} \sqrt{\cos y}}{\sqrt{\cos y} + \sqrt{\sin y}} - \int y \cdot \left( \frac{y^{2k+1} \sqrt{\cos y}}{\sqrt{\cos y} + \sqrt{\sin y}} \right)' dy$$

since we know $I_{n}$ for $2\:|\:n$. Still very complicated.

Vue
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