Pure · Integration · Year 2

Integrating Using Trigonometric Identities

No standard result integrates $\sin^2x$, $\cos^2x$ or $\tan^2x$ directly — rewrite them first using a double-angle identity, and the integral becomes routine.

9 min read 4 worked examples

Why identities are needed here

None of the standard results or the reverse chain rule can integrate $\sin^2x$: there's no function whose derivative is $\sin^2x$ sitting readily to hand. The fix is algebraic, not new calculus — use a trig identity to rewrite the integrand as something that is directly integrable, usually a constant plus a single sine or cosine term.

The double-angle rewrites

The double-angle identity $\cos2x = 1-2\sin^2x$ rearranges to give $\sin^2x$ in terms of $\cos2x$; the version $\cos2x=2\cos^2x-1$ does the same for $\cos^2x$.

Two identities worth memorising for integration

$\displaystyle\sin^2x = \dfrac{1-\cos2x}{2} \qquad\qquad \cos^2x = \dfrac{1+\cos2x}{2}$

Once rewritten this way, each term integrates using the standard results and the $f(ax+b)$ rule from earlier in this unit: the constant term gives a multiple of $x$, and the $\cos2x$ term gives a multiple of $\sin2x$.

Also useful

$\tan^2x = \sec^2x-1$, from rearranging $\sin^2x+\cos^2x=1$ (divide through by $\cos^2x$). This converts $\tan^2x$ — not directly integrable — into $\sec^2x-1$, which is.

Worked example 1

Integrating sin²x

Find $\displaystyle\int \sin^2x\,dx$.

MethodRewrite using the double-angle identity, then integrate term by term

$\sin^2x = \dfrac{1-\cos2x}{2}$.

Apply the double-angle rewrite — there's no direct standard result for sin²x.

$\displaystyle\int \sin^2x\,dx = \int \left(\dfrac12-\dfrac12\cos2x\right)dx$.

Split the fraction into two separately integrable terms.

$=\dfrac{x}2-\dfrac12\left(\dfrac12\sin2x\right)+c = \dfrac{x}2-\dfrac{\sin2x}4+c$.

Integrate the constant term directly, and cos(2x) using the f(ax+b) result with a = 2.

▪ solved

Worked example 2

Combining with f(ax + b)

Find $\displaystyle\int \cos^2(3x)\,dx$.

MethodApply the double-angle rewrite to the whole bracket 3x

$\cos^2(3x) = \dfrac{1+\cos(6x)}2$.

Apply the cos²θ rewrite with θ = 3x, so 2θ = 6x.

$\displaystyle\int \cos^2(3x)\,dx = \int \left(\dfrac12+\dfrac12\cos6x\right)dx$.

Split into two integrable terms.

$=\dfrac{x}2+\dfrac12\left(\dfrac16\sin6x\right)+c = \dfrac{x}2+\dfrac{\sin6x}{12}+c$.

Integrate the constant term directly, and cos(6x) using the f(ax+b) result with a = 6.

▪ solved

Worked example 3

Integrating tan²x

Find $\displaystyle\int \tan^2x\,dx$.

MethodRewrite tan²x using the Pythagorean identity, then integrate

$\tan^2x = \sec^2x-1$.

Apply the Pythagorean identity — tan²x has no standard antiderivative on its own.

$\displaystyle\int \tan^2x\,dx = \int \left(\sec^2x-1\right)dx$.

Rewrite the integral in terms of sec²x and a constant.

$= \tan x - x + c$.

Integrate sec²x directly using the standard result, and −1 using the power rule.

▪ solved

Worked example 4

A product using the sum formulas · Challenge

Find $\displaystyle\int \sin3x\cos x\,dx$.

MethodConvert the product into a sum using the compound-angle identities

$\sin A\cos B = \dfrac12\big[\sin(A+B)+\sin(A-B)\big]$.

This product-to-sum identity follows from adding the compound-angle expansions of sin(A+B) and sin(A−B).

With $A=3x$, $B=x$: $\sin3x\cos x = \dfrac12\big[\sin4x+\sin2x\big]$.

Substitute A = 3x, B = x.

$\displaystyle\int \sin3x\cos x\,dx = \dfrac12\int\left(\sin4x+\sin2x\right)dx$.

The integrand is now a sum of two integrable sine terms.

$= \dfrac12\left(-\dfrac{\cos4x}4-\dfrac{\cos2x}2\right)+c = -\dfrac{\cos4x}8-\dfrac{\cos2x}4+c$.

Integrate each term using the f(ax+b) result, with a = 4 and a = 2 respectively.

▪ solved

A product no other technique reaches

Neither the reverse chain rule nor a direct double-angle rewrite applies to a product of two different-angle trig functions like $\sin3x\cos x$ — the sum-formula rewrite is the only route in, converting a product into a sum of two separately integrable terms.

Common mistake

Trying to integrate $\sin^2x$ or $\cos^2x$ directly with the power rule, treating them like $x^2$. Powers of trig functions don't obey the power rule — the double-angle rewrite has to come first, every time, before any integration happens.

Check yourself

Find $\displaystyle\int \cos^2x\,dx$.

Method

$\cos^2x = \dfrac{1+\cos2x}2$.

Apply the double-angle rewrite.

$\displaystyle\int \cos^2x\,dx = \int\left(\dfrac12+\dfrac12\cos2x\right)dx = \dfrac{x}2+\dfrac{\sin2x}4+c$.

Integrate each term, using the f(ax+b) result on cos(2x).

▪ solved

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The Reverse Chain Rule

Integration by Substitution