Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
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) – user39572 Sep 10 '12 at 3:39
up vote 3 down vote accepted

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 $$\int(1-\cos^2 x) \ \cos^{n-2} x \ dx =\int \cos^{n-2} x \ dx-\int \cos^2 x\; \cos^{n-2} x \ dx \\=\int \cos^{n-2} x \ dx-\int \cos^{n} x \ dx $$

so we get

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

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


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

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

share|cite|improve this answer
Ah, just what I was looking for. Thanks Peter. – 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.

share|cite|improve this answer
Nice idea, thanks. – Joe Sep 10 '12 at 3:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.