Integrals · Lesson 3 of 9

Integrating with Trigonometric Identities

“Use identities to reveal standard integrals hidden inside trigonometric powers, products, and quotients.”

Learning Objectives

• Use power-reduction and product-to-sum identities to simplify integrands. • Choose between an identity and a substitution for trigonometric powers. • Simplify trigonometric quotients while retaining the original domain. • Explain why different-looking antiderivatives can be equivalent. • Use the principal-value interval when simplifying inverse trigonometric expressions.

An identity changes the form, not the function

An integral can look unfamiliar even though an equivalent integrand is easy to integrate. Trigonometric identities let us change the form while keeping the same value at every point where the expressions exist. The aim is to expose a sum of standard integrands or a function paired with its derivative.

Substitution and identities often work together. An odd power of sine can leave one sine factor for the differential while the remaining even power is written using cosine. When both powers are even, identities involving double angles often make a better starting point.

Reducing a squareLaTeX

sin^{2} u = \frac{1 - cos 2 u}{2}, cos^{2} u = \frac{1 + cos 2 u}{2}

These follow by rearranging the double-angle identity

cos 2 u = 1 - 2 sin^{2} u = 2 cos^{2} u - 1

. The angle

u

may itself depend on

x

Example — A squared cosine

Problem
Evaluate $\int cos^{2} x d x$ .

1.Replace the square by $(1 + cos 2 x) /2$ . This creates a constant term and a cosine term.
2.Integrate: $\frac{1}{2} \int 1 d x + \frac{1}{2} \int cos 2 x d x = x /2 + sin 2 x /4 + C$ .
3.The factor $1/4$ contains both the identity's factor $1/2$ and the chain-rule correction $1/2$ . Differentiating returns $(1 + cos 2 x) /2 = cos^{2} x$ .

Example — A fourth power

Problem
Evaluate $\int sin^{4} x d x$ .

1.First square the power-reduction identity: $sin^{4} x = (1 - 2 cos 2 x + cos^{2} 2 x) /4$ .
2.Reduce the remaining square using $cos^{2} 2 x = (1 + cos 4 x) /2$ . Collecting terms gives $3/8 - (1/2) cos 2 x + (1/8) cos 4 x$ .
3.Integrate each term, including its frequency factor: $3 x /8 - sin 2 x /4 + sin 4 x /32 + C$ .

Turning products into sums

A product such as $sin 2 x cos 3 x$ is not handled by multiplying two antiderivatives. Instead, the angle-addition identities turn it into a sum of functions that can be integrated separately. Carefully retain the sign when a difference of angles is negative.

Product-to-sum identitiesLaTeX

sin A cos B cos A cos B sin A sin B = \frac{1}{2} [sin (A + B) + sin (A - B)], = \frac{1}{2} [cos (A + B) + cos (A - B)], = \frac{1}{2} [cos (A - B) - cos (A + B)] .

Here

A

and

B

denote angles, not constants of integration. The first identity follows by adding the formulas for

sin (A + B)

and

sin (A - B)

; the others follow from cosine addition formulas.

Example — Different frequencies

Problem
Evaluate $\int sin 2 x cos 3 x d x$ .

1.Use the first identity: $sin 2 x cos 3 x = \frac{1}{2} [sin 5 x + sin (- x)] = \frac{1}{2} (sin 5 x - sin x)$ .
2.Integrate the two terms to obtain $- cos 5 x /10 + cos x /2 + C$ .
3.The negative-angle step is essential. Replacing $sin (- x)$ by $sin x$ would change the original integrand.

Example — A product of three cosines

Problem
How can $\int cos 2 x cos 4 x cos 6 x d x$ be simplified?

1.Combine the first two factors: $cos 2 x cos 4 x = \frac{1}{2} (cos 6 x + cos 2 x)$ .
2.Multiply by $cos 6 x$ . Reduce $cos^{2} 6 x$ and use product-to-sum on $cos 2 x cos 6 x$ .
3.The integrand becomes $\frac{1}{4} [1 + cos 12 x + cos 8 x + cos 4 x]$ .
4.The integral is $x /4 + sin 12 x /48 + sin 8 x /32 + sin 4 x /16 + C$ . Repeated use of the same identity is enough; no new integration rule is needed.

Choosing a useful representation of an odd power

An odd power can often be treated in more than one way. Comparing two correct solutions helps distinguish a genuine error from a harmless difference in appearance. The decisive test is whether the derivatives agree on the interval under consideration.

Example — Two answers for a cubed sine

Problem
Evaluate $\int sin^{3} x d x$ in two ways.

1.The triple-angle identity gives $sin^{3} x = (3 sin x - sin 3 x) /4$ . Integrating gives $- 3 cos x /4 + cos 3 x /12 + C$ .
2.Alternatively, write $sin^{3} x = (1 - cos^{2} x) sin x$ . Put $u = cos x$ and $d u = - sin x d x$ to get $- u + u^{3} /3 + C$ .
3.The second expression is $- cos x + cos^{3} x /3 + C$ . Substituting $cos 3 x = 4 cos^{3} x - 3 cos x$ into the first gives exactly the second expression.
4.More generally, antiderivatives with the same derivative can differ by a constant even if an identity does not make them look immediately identical.

Quotients and principal values

Quotients often simplify through double-angle or Pythagorean identities. Simplifying first can expose tangent, secant, or a denominator derivative. At the same time, cancellation must not cause us to forget excluded points, and inverse trigonometric functions must keep their specified output intervals.

Example — A squared tangent

Problem
Evaluate $\int tan^{4} x d x$ .

1.Write $tan^{4} x = tan^{2} x (sec^{2} x - 1) = tan^{2} x sec^{2} x - tan^{2} x$ .
2.In the first term set $u = tan x$ , giving $tan^{3} x /3$ . In the second use $tan^{2} x = sec^{2} x - 1$ , whose integral is $tan x - x$ .
3.Subtract to get $tan^{3} x /3 - tan x + x + C$ , on intervals where tangent exists.

Example — A quotient with a hidden square

Problem
Evaluate $\int (cos x - sin x) / (1 + sin 2 x) d x$ .

1.Since $1 + sin 2 x = (sin x + cos x)^{2}$ , rewrite the denominator in that form.
2.Put $u = sin x + cos x$ , so $d u = (cos x - sin x) d x$ .
3.The integral is $\int u^{- 2} d u = - u^{- 1} + C = - 1/ (sin x + cos x) + C$ .

Example — An inverse sine needs an interval

Problem
Evaluate $\int sin^{- 1} (cos x) d x$ on $0 < x < π$ .

1.Write $cos x = sin (π /2 - x)$ . On the given interval, $π /2 - x$ lies in $(- π /2, π /2)$ , the principal-value interval of inverse sine.
2.Therefore $sin^{- 1} (cos x) = π /2 - x$ on this interval.
3.Integrate to obtain $π x /2 - x^{2} /2 + C$ . Outside this interval the simplification must be reconsidered; it is not an identity valid for all real $x$ .

Common mistake

In general, $\int cos^{2} x d x$ is not $sin^{2} x /2$ , because differentiating that candidate gives $sin x cos x$ . Similarly, $sin^{- 1} (sin u) = u$ requires $u$ to lie in the principal-value interval. Identities and inverse-function simplifications have different domain conditions.

Quick check

Which identity is correct?

Quick check

What is $\int cos^{2} x d x$ ?

Quick check

Which expression equals $sin 2 x cos 3 x$ ?

Quick check

Which substitution naturally handles $sin^{3} x d x$ after reserving one sine factor?

Quick check

On $0 < x < π$ , what is $sin^{- 1} (cos x)$ ?

Practice Problems

Find $\int sin^{2} (2 x + 5) d x$ .
Find $\int sin 3 x cos 4 x d x$ .
Find $\int sin 4 x sin 8 x d x$ .
Find $\int cos^{4} (2 x) d x$ .
Find $\int sin^{3} x cos^{3} x d x$ and explain your choice of substitution.
Find $\int cos 2 x / (cos x + sin x)^{2} d x$ .
Find $\int tan^{3} (2 x) sec (2 x) d x$ .
Find $\int d x / [cos (x - a) cos (x - b)]$ when $sin (a - b) \neq = 0$ . First relate the integrand to $tan (x - a) - tan (x - b)$ .

Key Takeaways

Identities expose standard integrals without changing the integrand on its domain. Reduce even powers, convert products to sums, or reserve a factor for substitution. Retain chain-rule factors when integrating multiple angles. Different-looking antiderivatives can be equivalent. Inverse-trigonometric simplifications require principal-value and interval checks.

Previous · Lesson 2

Integration by Substitution

Next · Lesson 4

Standard Integrals Involving Quadratic Expressions

An identity changes the form, not the function

Turning products into sums

Choosing a useful representation of an odd power

Quotients and principal values

Quiz

Practice Problems

Key Takeaways