# Trigonometric Integrals Reference

## Basic Integrals

$$\int \sin x\, dx = -\cos x + c$$

$$\int \cos x\, dx = \sin x + c$$

$$\int \tan x\, dx = -\ln(\cos x) + c$$

$$\int \cot x\, dx = \ln(\sin x) + c$$

$$\int \sec x\, dx = \ln(\sec x + \tan x) + c$$

$$\int \csc x\, dx = \ln(\csc x - \cot x) + c$$

## Power-Reduction Formulas

The power-reduction formulas are useful for integrating powers of the trigonometric functions. For the sine and cosine functions, however, the series representation are probably easier because the power-reduction formulas require multiple integration steps until the power is reduced to the base power.

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

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

$$\int \tan^{n} x \, dx = \frac{1}{n-1}(\tan x)^{n-1} - \int (\tan x)^{n-2} \, dx$$

$$\int \cot^{n} x \, dx = -\frac{1}{n-1}(\cot x)^{n-1} - \int (\cot x)^{n-2} \, dx$$

$$\int \sec^{n} x \, dx = \frac{1}{n-1}\tan x(\sec x)^{n-2} + \frac{n-2}{n-1}\int (\sec x)^{n-2} \, dx$$

$$\int \csc^{n} x \, dx = -\frac{1}{n-1}\cot x(\csc x)^{n-2} + \frac{n-2}{n-1}\int (\csc x)^{n-2} \, dx$$

## Series Representation of Integrals

The series representation of integrals gives us a quick method of integrating the powers of trigonometric functions. The terms in the series terminate to finite terms.

$$\int (\cos x)^{2n+1} \, dx = \sin x - \frac{n}{3}\sin^3 x + \frac{n(n-1)}{5\cdot 2!}\sin^5 x - \frac{n(n-1)(n-2)}{7\cdot 3!}\sin^7 x + ...$$

A better series for cosine function for any power is given next. You can compare them and see that they both give the same integral.

$$\int (\cos x)^{n+1} \, dx =$$ $$\sin x + \frac{n}{2\cdot 3}\sin^3 x +$$ $$\frac{n(n-2)}{2^2\cdot 2!\cdot 5}\sin^5 x +$$ $$\frac{n(n-2)(n-4)}{2^3\cdot 3!\cdot 7}\sin^7 x + ...$$

$$\int (\tan x)^{2n+1} \, dx = (-1)^n\left[ \ln(\sec x) - \frac{n}{2}\sec^2 x + \frac{n(n-1)}{4\cdot 2!}\sec^4 x - \frac{n(n-1)(n-2)}{6\cdot 3!}\sec^6 x + ... \right]$$

$$\int (\cot x)^{2n+1} \, dx = (-1)^n\left[ -\ln(\csc x) + \frac{n}{2}\csc^2 x - \frac{n(n-1)}{4\cdot 2!}\csc^4 x + \frac{n(n-1)(n-2)}{6\cdot 3!}\csc^6 x + ... \right]$$

$$\int (\sec x)^{2n+2} \, dx = \tan x + \frac{n}{3}\tan^3 x + \frac{n(n-1)}{5\cdot 2!}\tan^5 x + \frac{n(n-1)(n-2)}{7\cdot 3!}\tan^7 x + ...$$

$$\int (\csc x)^{2n+2} \, dx = -\cot x - \frac{n}{3}\cot^3 x - \frac{n(n-1)}{5\cdot 2!}\cot^5 x - \frac{n(n-1)(n-2)}{7\cdot 3!}\cot^7 x + ...$$

## Pascal’s Triangle in Trigonometric Integrals

$$\int \cos x \, dx = \sin x + c$$

$$\int \cos^3 x \, dx = \sin x - \frac{1}{3}\sin^3 x + c$$

$$\int \cos^5 x \, dx = \sin x - \frac{2}{3}\sin^3 x + \frac{1}{5}\sin^5 x + c$$

$$\int \cos^7 x \, dx = \sin x - \frac{3}{3}\sin^3 x + \frac{3}{5}\sin^5 x - \frac{1}{7}\sin^7 x + c$$

$$\int \cos^9 x \, dx = \sin x - \frac{4}{3}\sin^3 x + \frac{6}{5}\sin^5 x - \frac{4}{7}\sin^7 x + \frac{1}{9}\sin^9 x + c$$

The general formula is:

$$\int (\cos x)^{2n+1} \, dx = \sum_{i=1}^{n} \frac{(-1)^{i}C_{i}^{n}(\sin x)^{2i+1}}{2i+1}$$ for all integers n, where $$C_{m}^{n} = \frac{n!}{m!(n-m)!}$$

$$\int \sin x \, dx = -\cos x + c$$

$$\int \sin^3 x \, dx = -\cos x + \frac{1}{3}\cos^3 x + c$$

$$\int \sin^5 x \, dx = -\cos x + \frac{2}{3}\cos^3 x - \frac{1}{5}\cos^5 x + c$$

$$\int \sin^7 x \, dx = -\cos x + \frac{3}{3}\cos^3 x - \frac{3}{5}\cos^5 x + \frac{1}{7}\cos^7 x + c$$

$$\int \sin^9 x \, dx = -\cos x + \frac{4}{3}\cos^3 x - \frac{6}{5}\cos^5 x + \frac{4}{7}\cos^7 x - \frac{1}{9}\cos^9 x + c$$

## Proofs

Proof of the cosine series:

(i) $$\cos^n x = (1-\sin^2 x)^{n/2} =$$ $$1 + \frac{n}{2}\sin^2 x + \frac{(n/2)(n/2-1)}{2!}\sin^4 x +$$ $$\frac{(n/2)(n/2-1)(n/2-2)}{3!}\sin^6 x + ...$$  (Binomial expansion)

(ii) $$\cos^n x = 1 + \frac{n}{2}\sin^2 x +$$ $$\frac{n(n-2)}{2^2\cdot 2!}\sin^4 x +$$ $$\frac{n(n-2)(n-4)}{2^3\cdot 3!}\sin^6 x + ...$$

(iii) $$\cos^n x\cdot \cos x =$$ $$\cos x + \frac{n}{2}\sin^2 x\cdot \cos x +$$ $$\frac{n(n-2)}{2^2\cdot 2!}\sin^4 x\cdot \cos x +$$ $$\frac{n(n-2)(n-4)}{2^3\cdot 3!}\sin^6 x\cdot \cos x + ...$$   (Multiply both sides by $$\cos x$$)

(iv) $$\int (\cos x)^{n+1} \, dx =$$ $$\sin x + \frac{n}{2\cdot 3}\sin^3 x +$$ $$\frac{n(n-2)}{2^2\cdot 2!\cdot 5}\sin^5 x +$$ $$\frac{n(n-2)(n-4)}{2^3\cdot 3!\cdot 7}\sin^7 x + ...$$

## Alternate Cosine Power Integrals

An alternate way of integrating the powers of cosine is to write the powers of cosine in terms of the multiple-angle identities. Solving for the powers of cosines from the identities leads to the identities of the powers of cosines.

$$\cos^2 x = \frac{1}{2}\left[ \cos(2x) + 1 \right]$$

$$\cos^3 x = \frac{1}{2^2}\left[ \cos(3x) + 3\cos x \right]$$

$$\cos^4 x = \frac{1}{2^3}\left[ \cos(4x) + 4\cos(2x) + 3 \right]$$

$$\cos^5 x = \frac{1}{2^4}\left[ \cos(5x) + 5\cos(3x) + 10\cos x \right]$$

$$\cos^6 x = \frac{1}{2^5}\left[ \cos(6x) + 6\cos(4x) + 15\cos(2x) + 10 \right]$$

$$\cos^7 x = \frac{1}{2^6}\left[ \cos(7x) + 7\cos(5x) + 21\cos(3x) + 35\cos x \right]$$

$$\cos^8 x = \frac{1}{2^7}\left[ \cos(8x) + 8\cos(6x) + 28\cos(4x) + 56\cos(2x) + 35 \right]$$

The pattern reveals the multiple n of angle x in cos(nx) terms all decrease by 2. Their coefficients are from the first half of Pascal’s Triangle. In even powers, however, the constant is from the preceding row.