Prove that $(2n+1)k_{n+1}=(2n+1)k_{n}+\cos^{2n+1} (x)$ Given that $$k_n=\int \frac{\cos^{2n} (x)}{\sin (x)} dx$$ Prove that $$(2n+1)k_{n+1}=(2n+1)k_{n}+\cos^{2n+1} (x)$$
I have tried to prove this is true by differentiating both sides with product rule:
$$2k_{n+1}+\frac{\cos^{2n+1} (x)}{\sin (x)}(2n+1)=2k_n+\frac{\cos^{2n} (x)}{\sin (x)}(2n+1)+(2n+1)\cos^{2n} (x) \sin (x)$$
I am stuck here as I met a dead end upon grouping and expanding. Please help. Thank you in advance!!

Verifying by differentiation is a good idea. We wish to prove that
$$k_{n+1}=k_n+\frac{1}{2n+1}\cos^{2n+1}(x).$$
Recall that indefinite integrals are only determined up to a constant. So we really need to show that the derivative of the left-hand side is equal to the derivative of the right-hand side. By the Fundamental Theorem of Calculus, we need to show therefore that
$$\frac{\cos^{2n+2}(x)}{\sin x}=\frac{\cos^{2n}(x)}{\sin x}-\sin x\cos^{2n}(x).$$
Equivalently, we want to show that
$$\sin x\cos^{2n}(x)=\frac{\cos^{2n}(x)}{\sin x}-\frac{\cos^{2n+2}(x)}{\sin x}.\tag{1}$$
The right-hand side of (1) is $\frac{\cos^{2n}(x)}{\sin x}(1-\cos^2 x)$. Replace $1-\cos^2 x$ by $\sin^2 x$, cancel a $\sin x$ from top and bottom, and we obtain the left-hand side of (1).
