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!!
 A: We have 
$$\begin{align}
k_{n+1}&=\int \frac{\cos^{2n+2 }x}{\sin x}\,dx\\\\
&=\int \frac{(1-\sin^2 x)\cos^{2n}x}{\sin x}\,dx\\\\
&=\int \frac{\cos^{2n x}}{\sin x}\,dx-\int \sin x\cos ^{2n}x\,dx\\\\
&=k_n+\frac{\cos^{2n+1}x}{2n+1}
\end{align}$$
Therefore, we have
$$\bbox[5px,border:2px solid #C0A000]{(2n+1)k_{n+1}=(2n+1)k_n+\cos^{2n+1}x}$$
A: Let $t = \cos x$, then $$k_n = \int \frac{t^{2n}}{t^2-1}dt.$$
One has $$k_{n+1} = \int \frac{t^{2n+2}}{t^2-1}dt = \int \frac{t^{2n}(t^2-1+1)}{t^2-1}dt = \int t^{2n}dt + k_n = \frac{t^{2n+1}}{2n+1} + k_n.$$
Then, you got the conclusion.
A: 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).
