Trigonometric identities that make a substitution possible

What you will be able to do

• integrate $\sin^2 x$ and $\cos^2 x$ with the half-angle identities;
• integrate odd powers of sine and cosine by peeling off one factor;
• turn a product of a sine and a cosine into a sum, and integrate it.

The idea

An integral such as $\int \sin^2 x\,dx$ has no inner function with its derivative beside it, so plain substitution fails. But $\sin^2 x$ can be rewritten as a sum of simple terms. The rewriting is an identity, and the integral is then easy.

Squares of sine and cosine

The double angle formula $\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1$ gives

So $\displaystyle\int \cos^2 x\,dx = \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$. The term $\dfrac{x}{2}$ appears because the average height of $\cos^2 x$ is $\dfrac{1}{2}$, and the wavy part has no lasting growth.

Odd powers: peel off one factor

When the power of $\sin x$ is odd, keep one $\sin x$ for $du$, and write the rest with $\cos x$. The same works for an odd power of $\cos x$.

Example. Find $\displaystyle\int \sin^3 x\,dx$.

1. Write $\sin^3 x = \sin x\,(1 - \cos^2 x)$.
2. Let $u = \cos x$, so $du = -\sin x\,dx$.
3. The integral becomes $-\int (1 - u^2)\,du = -u + \dfrac{u^3}{3} + C$.
4. Return to $x$: $-\cos x + \dfrac{\cos^3 x}{3} + C$.

Example. Find $\displaystyle\int \sin^2 x\cos^3 x\,dx$. The power of $\cos x$ is odd.

1. Write $\cos^3 x = \cos x\,(1 - \sin^2 x)$ and let $u = \sin x$, so $du = \cos x\,dx$.
2. The integral becomes $\int u^2(1 - u^2)\,du = \dfrac{u^3}{3} - \dfrac{u^5}{5} + C$.
3. Return to $x$: $\dfrac{\sin^3 x}{3} - \dfrac{\sin^5 x}{5} + C$.

If both powers are even, use the half-angle identities instead.

Products of sines and cosines

The product formulas come from adding the formulas for $\sin(A + B)$ and $\sin(A - B)$.

• $2\sin A\cos B = \sin(A + B) + \sin(A - B)$
• $2\cos A\cos B = \cos(A + B) + \cos(A - B)$
• $2\sin A\sin B = \cos(A - B) - \cos(A + B)$

Example. Find $\displaystyle\int \sin 3x\cos 2x\,dx$.

1. Use the first formula: $\sin 3x\cos 2x = \tfrac{1}{2}\left(\sin 5x + \sin x\right)$.
2. Integrate each term: $\tfrac{1}{2}\left(-\dfrac{\cos 5x}{5} - \cos x\right) + C$.
3. The answer is $-\dfrac{\cos 5x}{10} - \dfrac{\cos x}{2} + C$.

Powers of tangent and secant

The identity $\tan^2 x = \sec^2 x - 1$ turns $\int \tan^2 x\,dx$ into $\int (\sec^2 x - 1)\,dx = \tan x - x + C$.

For $\int \sec^4 x\,dx$ write $\sec^4 x = \sec^2 x\,(1 + \tan^2 x)$ and let $u = \tan x$. Then $du = \sec^2 x\,dx$, and the integral is $\tan x + \dfrac{\tan^3 x}{3} + C$.

A last identity: $1 + \cos x = 2\cos^2\dfrac{x}{2}$. So $\displaystyle\int \dfrac{dx}{1 + \cos x} = \int \tfrac{1}{2}\sec^2\dfrac{x}{2}\,dx = \tan\dfrac{x}{2} + C$.

Practise with new numbers

Try it in Python

The program first checks the identity for $\sin^2 x$ at three values of $x$. Then it tests two antiderivatives that came from identities, by comparing slopes.

Common mistakes

• Writing $\sin^2 x$ as $\sin(x^2)$. They are different functions. The square belongs to the value of $\sin x$.
• Losing the factor from the angle. $\int \cos 2x\,dx = \dfrac{\sin 2x}{2}$, not $\sin 2x$.
• Wrong sign. $\int \sin x\,dx = -\cos x$, and $du = -\sin x\,dx$ when $u = \cos x$.
• Using a half-angle identity for an odd power. Peeling off one factor is shorter and safer.
• Not checking. Differentiate the answer. A wrong sign or a wrong factor shows up at once.

Remember these

• $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ and $\cos^2 x = \dfrac{1 + \cos 2x}{2}$.
• For an odd power of $\sin x$ or $\cos x$, keep one factor for $du$ and use $\sin^2 x + \cos^2 x = 1$.
• $2\sin A\cos B = \sin(A + B) + \sin(A - B)$ turns a product into a sum.
• $\tan^2 x = \sec^2 x - 1$ and $\sec^2 x = 1 + \tan^2 x$ help with powers of tangent and secant.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

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

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

3. Find $\displaystyle\int \sin 2x\cos 4x\,dx$.

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

5. Find $\displaystyle\int \dfrac{dx}{1 - \cos x}$.

References

• OpenStax, Calculus Volume 2, Chapter 3, section 3.2 Trigonometric Integrals.
• NCERT, Mathematics Part II for Class XII, Chapter 7, the section on integration using trigonometric identities.