I am trying to prove $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$

This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.

$$I_n := \int\cos^n x \ dx = \int\cos^{n-1} x \cos x \ dx \tag{1}$$

First question: why rewrite the original instead of immediately integrating by parts of $\int \cos^n x \ dx$?

Integrate by parts with $$u = \cos^{n-1} x, dv = \cos x \ dx \implies du = (n-1)\cos^{n-2} x \cdot -\sin x, v = \sin x$$

which leads to

$$I_n = \sin x \ \cos^{n-1} x +\int\sin^2 x (n-1) \ \cos^{n-2} x \ dx \tag{2}$$

Since $(n-1)$ is a constant, we can throw it out front of the integral:

$$I_n = \sin x \ \cos^{n-1} x +(n-1)\int\sin^2 x \ \cos^{n-2} x \ dx\tag{3}$$

I can transform the integral a bit because $\sin^2 x + cos^2 x = 1 \implies \sin^2 x = 1-\cos^2 x$

$$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$

According to Wikipedia as noted here, this simplifies to:

$$I_n = \sin x \ \cos^{n-1} x + (n-1) \int \cos^{n-2} x \ dx - (n-1)\int(\cos^n x) \ dx \tag{5}$$

Question 2: How did they simplify the integral of $\int(1-\cos^2 x) \ dx$ to $\int(\cos^n x) \ dx$?

Assuming knowledge of equation 5, I see how to rewrite it as

$$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x - (n-1) I_{n} \tag{6}$$

and solve for $I_n$. I had tried exploiting the fact that $$\cos^2 x = \frac{1}{2} \cos(2x) + \frac{1}{2} $$

and trying to deal with $\int 1 \ dx - \int \frac{1}{2} \cos (2x) + \frac{1}{2} \ dx$

which left me with $\frac{x}{2} - \frac{1}{4} \sin(2x)$ after integrating those pieces. Putting it all together I have:

$$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x \left(-(n-1) (\frac{x}{2} - \frac{1}{4} \sin 2x) \right) \tag{7}$$

but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of

$$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$

  • 2
    $\begingroup$ Q1: re-writing it as such is simply a way to make the IBP clear. Q2: $(1-\cos^2x)\cos^{n-2}=\cos^{n-2}- \cos^n x$ (the product is simply expanded, since it is easier to deal with the sum of two simple integrals than the integral of a product) $\endgroup$ – user39572 Sep 10 '12 at 3:39
  • $\begingroup$ math.stackexchange.com/questions/3168506/… $\endgroup$ – lab bhattacharjee Mar 31 '19 at 10:07

I guess your problem is step four:

$$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$

Note that $$\begin{align}\int(1-\cos^2 x) \ \cos^{n-2} x \ dx & = \int \cos^{n-2} x \ dx-\int \cos^2 x\; \cos^{n-2} x \ dx \\ & =\int \cos^{n-2} x \ dx-\int \cos^{n} x \ dx \end{align}$$

so we get

$$\begin{align}I_n & = \sin x \ \cos^{n-1} x + (n-1)\int \cos^{n-2} x \ dx-(n-1)\int \cos^{n} x \ dx \\ I_n & = \sin x \ \cos^{n-1} x + (n-1)I_{n-2}-(n-1)I_n \end{align}$$


$$nI_n = \sin x \ \cos^{n-1} x + (n-1)I_{n-2}$$

$$I_n =\frac{ \sin x \ \cos^{n-1} x}n + \frac{n-1}nI_{n-2}$$

  • $\begingroup$ Ah, just what I was looking for. Thanks Peter. $\endgroup$ – Joe Sep 10 '12 at 3:58

How about verifying using differentiation? We are asked to show $$\int_0^x \cos^n t dt = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n} \int^x_0 \cos^{n-2} t dt + C$$ for some constant $C$. This is true iff the derivatives of each side are equal.

Differentiating the right side, we get using the product rule $$RHS = \frac{n-1}{n} \cos^{n-2} x (-\sin x) \sin x + \frac{1}{n} \cos^{n-1} x \cos x + \frac{n-1}{n} \cos^{n-2} x,$$ then combining, $$RHS = \frac{n-1}{n} (1-\sin^2 x) \cos^{n-2} x + \frac{1}{n} \cos^n x$$ and finally using $\cos^2 x + \sin^2 x = 1$, $$RHS = \frac{n-1}{n}\cos^n x+\frac{1}{n} \cos^nx = \cos^n x,$$ which is the what we want.

  • $\begingroup$ Nice idea, thanks. $\endgroup$ – Joe Sep 10 '12 at 3:58

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