Mathematics 9709 · Pure Mathematics 2

Further integration of trigonometric functions

Powers like $\sin^2x$ or $\tan^2x$ won't integrate on their own.
The fix is always the same one move: rewrite the power with an identity until every term is a plain $\sin$ , $\cos$ or constant you can integrate from §5.3.
The whole skill is spotting which identity to use.

A squared trig function tells you to use an identity: $\sin^2x=\tfrac12(1-\cos2x)$ , $\cos^2x=\tfrac12(1+\cos2x)$ , $\tan^2x=\sec^2x-1$ . The square goes away and you integrate one term at a time.

The rewrites

Double-angle swap: $\sin^2x=\tfrac12(1-\cos2x)$ and $\cos^2x=\tfrac12(1+\cos2x)$ .
$\tan^2x=\sec^2x-1$ (which gives $\int\tan^2x\,dx=\tan x-x+c$ ).
Products like $\sin3x\sin x$ : break them up with $\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]$ .

Why you can't integrate a squared trig term as it is

Integration reverses differentiation. But nothing we know differentiates straight to $\sin^2x$ . The chain and product rules always give a $\sin x\cos x$ term, never a plain square. The double-angle identity fixes this. It swaps the square you can't integrate for $\tfrac12-\tfrac12\cos2x$ , which is two things you can reverse. So “see a square, swap it for an identity” is the only way to do it.

Worked examples

Worked example

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

Swap the square for the identity: $\sin^2x=\tfrac12(1-\cos2x)$ .
Integrate term by term — the $\cos2x$ carries a $\tfrac12$ from $a=2$ : $\int\tfrac12(1-\cos2x)\,dx=\tfrac12\Big(x-\tfrac12\sin2x\Big)+c$ .
Tidy: $\tfrac12 x-\tfrac14\sin2x+c$ .

Answer

\tfrac12 x-\tfrac14\sin2x+c

Worked example

Real question: $\int_{\pi/6}^{\pi/3}\dfrac{1-\cos 2\theta}{1+\cos 2\theta}\,d\theta$ (after proving it equals $\tan^2\theta$ )

9709/31 May/June 2021 Q4. Part (a) proves $\dfrac{1-\cos2\theta}{1+\cos2\theta}\equiv\tan^2\theta$ ; part (b) is the integral that this rewrite then lets you do.

Use part (a) to replace the fraction with $\tan^2\theta$ , then the identity $\tan^2\theta=\sec^2\theta-1$ : $\int_{\pi/6}^{\pi/3}\!\big(\sec^2\theta-1\big)\,d\theta$ .
Integrate: $\sec^2\theta\to\tan\theta$ and $1\to\theta$ , giving $\big[\tan\theta-\theta\big]_{\pi/6}^{\pi/3}$ .
Top limit: $\tan\tfrac{\pi}{3}-\tfrac{\pi}{3}=\sqrt3-\tfrac{\pi}{3}$ . Bottom: $\tan\tfrac{\pi}{6}-\tfrac{\pi}{6}=\tfrac{1}{\sqrt3}-\tfrac{\pi}{6}$ .
Subtract: $\sqrt3-\tfrac{1}{\sqrt3}-\big(\tfrac{\pi}{3}-\tfrac{\pi}{6}\big)=\tfrac{2}{\sqrt3}-\tfrac{\pi}{6}$ , since $\sqrt3-\tfrac{1}{\sqrt3}=\tfrac{2}{\sqrt3}$ .

Answer

\tfrac{2\sqrt3}{3}-\tfrac{\pi}{6}

Where the marks go

Common mistake

Keep the

\tfrac12

from the identity:

\sin^2x=\tfrac12(1-\cos2x)

, not

1-\cos2x

. And every

\cos(kx)

still gets its own

\tfrac1k

when you integrate.

Common mistake

For

\tan^2

, use

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

, not the double-angle identity. Picking the wrong identity is the usual mistake here. (9709/31 May/June 2021 Q4)

Now you try

Given that $\cos^4\theta-\sin^4\theta+4\sin^2\theta\cos^2\theta$ tidies up to $\cos2\theta+\sin^22\theta$ , find the exact value of $\int_{-\pi/18}^{\pi/18}\!\big(\cos^4\theta-\sin^4\theta+4\sin^2\theta\cos^2\theta\big)\,d\theta$ . (9709/32 May/June 2024 Q7)

Your turn— tap to reveal the worked answer (9709/32 May/June 2024 Q7)

The thing you're integrating is $\cos2\theta+\sin^22\theta$ . Swap the square again with $\sin^22\theta=\tfrac12(1-\cos4\theta)$ , so it becomes $\tfrac12+\cos2\theta-\tfrac12\cos4\theta$ .

Integrate: $\big[\tfrac12\theta+\tfrac12\sin2\theta-\tfrac18\sin4\theta\big]_{-\pi/18}^{\pi/18}$ . The limits are equal and opposite, so each term just doubles its top value:

Answer

\tfrac{\pi}{18}+\sin\tfrac{\pi}{9}-\tfrac14\sin\tfrac{2\pi}{9}

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Further integration of trigonometric functions

Powers like $\sin^2x$ or $\tan^2x$ won't integrate on their own.
The fix is always the same one move: rewrite the power with an identity until every term is a plain $\sin$ , $\cos$ or constant you can integrate from §5.3.
The whole skill is spotting which identity to use.

The rewrites

Double-angle swap: $\sin^2x=\tfrac12(1-\cos2x)$ and $\cos^2x=\tfrac12(1+\cos2x)$ .
$\tan^2x=\sec^2x-1$ (which gives $\int\tan^2x\,dx=\tan x-x+c$ ).
Products like $\sin3x\sin x$ : break them up with $\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]$ .

Why you can't integrate a squared trig term as it is

Worked examples

Worked example

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

Swap the square for the identity: $\sin^2x=\tfrac12(1-\cos2x)$ .
Integrate term by term — the $\cos2x$ carries a $\tfrac12$ from $a=2$ : $\int\tfrac12(1-\cos2x)\,dx=\tfrac12\Big(x-\tfrac12\sin2x\Big)+c$ .
Tidy: $\tfrac12 x-\tfrac14\sin2x+c$ .

Answer

\tfrac12 x-\tfrac14\sin2x+c

Worked example

Real question: $\int_{\pi/6}^{\pi/3}\dfrac{1-\cos 2\theta}{1+\cos 2\theta}\,d\theta$ (after proving it equals $\tan^2\theta$ )

9709/31 May/June 2021 Q4. Part (a) proves $\dfrac{1-\cos2\theta}{1+\cos2\theta}\equiv\tan^2\theta$ ; part (b) is the integral that this rewrite then lets you do.

Use part (a) to replace the fraction with $\tan^2\theta$ , then the identity $\tan^2\theta=\sec^2\theta-1$ : $\int_{\pi/6}^{\pi/3}\!\big(\sec^2\theta-1\big)\,d\theta$ .
Integrate: $\sec^2\theta\to\tan\theta$ and $1\to\theta$ , giving $\big[\tan\theta-\theta\big]_{\pi/6}^{\pi/3}$ .
Top limit: $\tan\tfrac{\pi}{3}-\tfrac{\pi}{3}=\sqrt3-\tfrac{\pi}{3}$ . Bottom: $\tan\tfrac{\pi}{6}-\tfrac{\pi}{6}=\tfrac{1}{\sqrt3}-\tfrac{\pi}{6}$ .
Subtract: $\sqrt3-\tfrac{1}{\sqrt3}-\big(\tfrac{\pi}{3}-\tfrac{\pi}{6}\big)=\tfrac{2}{\sqrt3}-\tfrac{\pi}{6}$ , since $\sqrt3-\tfrac{1}{\sqrt3}=\tfrac{2}{\sqrt3}$ .

Answer

\tfrac{2\sqrt3}{3}-\tfrac{\pi}{6}

Where the marks go

Common mistake

Keep the

\tfrac12

from the identity:

\sin^2x=\tfrac12(1-\cos2x)

, not

1-\cos2x

. And every

\cos(kx)

still gets its own

\tfrac1k

when you integrate.

Common mistake

For

\tan^2

, use

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

, not the double-angle identity. Picking the wrong identity is the usual mistake here. (9709/31 May/June 2021 Q4)

Now you try

Your turn— tap to reveal the worked answer (9709/32 May/June 2024 Q7)

The thing you're integrating is $\cos2\theta+\sin^22\theta$ . Swap the square again with $\sin^22\theta=\tfrac12(1-\cos4\theta)$ , so it becomes $\tfrac12+\cos2\theta-\tfrac12\cos4\theta$ .

Integrate: $\big[\tfrac12\theta+\tfrac12\sin2\theta-\tfrac18\sin4\theta\big]_{-\pi/18}^{\pi/18}$ . The limits are equal and opposite, so each term just doubles its top value:

Answer

\tfrac{\pi}{18}+\sin\tfrac{\pi}{9}-\tfrac14\sin\tfrac{2\pi}{9}