Looking for the recurrence relation for certain trigonometric integrals By assuming that:
$$ \int_{\pi/4}^{\pi/2} \frac{\cos^4(x)}{\sin^5(x)}\,dx = k,$$
what does the integral $$ \int_{\pi/4}^{\pi/2} \frac{\cos^6(x)}{\sin^7(x)}\,dx$$ equal in terms of k?
I have manipulated both integrals to get $\int_{\pi/4}^{\pi/2} (\csc^2(x)-1)^2(\csc(x))\,dx$ and $\int_{\pi/4}^{\pi/2}\csc^2(x)-1)^3(\csc(x))\,dx$.
I'm not sure where to go from here.
I multiplied $(\csc^2(x)-1)^2$ and $(\csc^2(x)-1)^3$ out, but I still couldn't find a solution. 
Please help! 
 A: By setting
$$ I_n = \int_{\pi/4}^{\pi/2}\frac{\cos^{2n}x}{\sin^{2n+1}x}dx$$
we have

$$ I_n = \int_{0}^{1}\frac{u^{2n}}{\sqrt{1+u^2}}du.\tag{1}$$

This follows from the variable changes $x=\arcsin t, t=\sqrt{s}, s=1/r, r=u+1$, or just $x=\arcsin\left(\frac{1}{\sqrt{1+u}}\right)$, for short.
Since:
$$\frac{d}{du}\operatorname{arcsinh} u = \frac{1}{\sqrt{1+u^2}},\qquad \frac{d}{du}\sqrt{1+u^2}=\frac{u}{\sqrt{1+u^2}},$$
integration by parts gives that $I_n$ is always a linear combination of $\sqrt{2}$ and $\operatorname{arcsinh}(1)=-\log\tan\frac{\pi}{8}=\log(1+\sqrt{2})$ with rational coefficients. In facts:
$$I_n + I_{n+1} = \int_{0}^{1}u^{2n}\sqrt{1+u^2}du=\frac{\sqrt{2}}{2n+1}-\frac{1}{2n+1}\cdot I_{n+1},$$
or:
$$I_{n+1} = \frac{\sqrt{2}-(2n+1)\cdot I_n}{2n+2},$$

$$ (2n+1)\cdot  I_n + (2n+2)\cdot I_{n+1} = \sqrt{2},$$
$$ I_0 = \log(1+\sqrt{2}).\tag{2}$$

We can also write $I_n$ as a value of the incomplete beta function:
$$I_n = \frac{1}{2}\cdot B\left(\frac{1}{2},n+\frac{1}{2},-n\right),$$
or recover $I_n$ from the coefficients of a Taylor series, since $(1)$ gives:

$$\sum_{n=0}^{+\infty}I_n\, x^{2n}=\frac{1}{\sqrt{1+x^2}}\cdot\operatorname{arctanh}\left(\sqrt{\frac{1+x^2}{2}}\right)=\frac{1}{2\sqrt{1+x^2}}\cdot\log\left(\frac{\sqrt{2}+\sqrt{1+x^2}}{\sqrt{2}-\sqrt{1+x^2}}\right).\tag{3}$$

Exploiting convexity we have:
$$I_n\geq \int_{1-\frac{2}{4n-1}}^{1}\frac{(4n-1)u-(4n-3)}{2\sqrt{2}}\,du = \frac{1}{\sqrt{2}(4n-1)}>\frac{1}{4\sqrt{2}\,n},$$
and by plugging this inequality into $(2)$ we get:
$$ I_n < \frac{3}{4\sqrt{2}\,n}.$$
Since $I_n = \frac{\sqrt{2}}{2n+1}-\frac{I_{n+1}}{2n+1}$, we get the asymptotic:

$$ I_n = \frac{1}{2\sqrt{2}\,n}+\Theta\left(\frac{1}{n^2}\right).\tag{4}$$

Integration by parts gives other terms of the asymptotics, in virtue of:
$$I_n -\frac{1}{(2n+1)\sqrt{2}}= \int_{0}^{1}\left(\frac{u^{2n}}{\sqrt{1+u^2}}-\frac{u^{2n}}{\sqrt{2}}\right)du=\int_{0}^{1}\frac{u^{2n}\left(\sqrt{2}+\sqrt{1+u^2}\right)}{(1-u^2)\sqrt{2(1+u^2)}}\,du.$$
