Prove that $(2n+1)k_{n+1}=(2n+1)k_{n}+\cos^{2n+1} (x)$

Question

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!!

Answer 1

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

$$\bbox[5px,border:2px solid #C0A000]{(2n+1)k_{n+1}=(2n+1)k_n+\cos^{2n+1}x}$$

Answer 2

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.

Answer 3

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).