Integral of sin x sin 2x (Calculus Guide)

To integrate (\sin x\sin 2x), rewrite the product with the identity (\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]). This gives (\tfrac12(\cos x-\cos 3x)). Integrate each cosine, including the factor (\tfrac13) for (\cos 3x), to obtain (\tfrac12\sin x-\tfrac16\sin 3x+C).

A product of two sine functions can be harder to integrate than either function alone. The key is to change the product into a sum, then use familiar cosine antiderivatives. I’ll show each step, explain the most common mistake, and check the result by differentiating it.

Integral Overview & Method Selection

The integral (\int\sin x\sin 2x\,dx) asks for a function whose derivative is the product (\sin x\sin 2x). A product-to-sum identity turns that product into two cosine terms, which are straightforward to integrate. This method also makes the chain-rule factor in the (3x) term clear.

The expression has two sine factors with different angles: (x) and (2x). A product-to-sum identity is a natural method because it converts a product of trigonometric functions into a difference of cosines. Once the expression is rewritten, the integration uses the standard cosine rule.

Here, (x) is understood to be measured in radians, as is standard in calculus. The notation (C) represents an arbitrary constant. It appears because the derivative of any constant is zero, so many functions can have the same derivative.

There is another valid route. Since (\sin 2x=2\sin x\cos x), the product becomes (2\sin^2x\cos x), which can be integrated using substitution. Both methods give equivalent answers. The product-to-sum method is especially useful when you are practicing trigonometric identities or working with more general products.

Trigonometric Identity and Integration Steps

The product-to-sum identity states that (\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]). For this integral, set (A=x) and (B=2x), then simplify the angles. Integrating the resulting cosine terms gives the antiderivative, provided the factor from the inner angle (3x) is included.

1. Choose the identity and substitute the angles

Start with the identity:

[ \sin A\sin B=\frac12\bigl[\cos(A-B)-\cos(A+B)\bigr]. ]

For the given product, (A=x) and (B=2x). Substitution gives

[ \sin x\sin 2x =\frac12\bigl[\cos(x-2x)-\cos(x+2x)\bigr]. ]

Now simplify the angle expressions:

[ \sin x\sin 2x =\frac12\bigl[\cos(-x)-\cos 3x\bigr]. ]

Cosine is an even function, so (\cos(-x)=\cos x). Therefore,

[ \sin x\sin 2x=\frac12(\cos x-\cos 3x). ]

The factor (\tfrac12) applies to both cosine terms. Keeping it visible helps prevent a sign or factor error in the next step.

2. Integrate each term

Replace the product in the integral with its cosine form:

[ \int\sin x\sin 2x\,dx =\frac12\int(\cos x-\cos 3x)\,dx. ]

Integrate term by term:

[ =\frac12\left(\int\cos x\,dx-\int\cos 3x\,dx\right). ]

The first antiderivative is (\sin x). For the second, the inner angle is (3x), so

[ \int\cos 3x\,dx=\frac13\sin 3x. ]

This factor can be checked by differentiating: the derivative of (\sin 3x) is (3\cos 3x), so multiplying by (\tfrac13) cancels the extra (3).

Substitute both antiderivatives:

[ \int\sin x\sin 2x\,dx =\frac12\left(\sin x-\frac13\sin 3x\right)+C. ]

Distribute the outer factor:

[ \boxed{\int\sin x\sin 2x\,dx =\frac12\sin x-\frac16\sin 3x+C.} ]

The coefficient of (\sin 3x) is (-\tfrac16), not (-\tfrac12). It includes both the outer factor (\tfrac12) and the integration factor (\tfrac13).

A second method using substitution

The double-angle identity gives (\sin 2x=2\sin x\cos x). Thus,

[ \int\sin x\sin 2x\,dx =\int 2\sin^2x\cos x\,dx. ]

Let (u=\sin x). Then (du=\cos x\,dx), so the integral becomes

[ \int 2u^2\,du=\frac23u^3+C =\frac23\sin^3x+C. ]

This looks different from the boxed answer, but it is the same antiderivative in another form. Using (\sin 3x=3\sin x-4\sin^3x),

[ \frac12\sin x-\frac16\sin 3x =\frac12\sin x-\frac16(3\sin x-4\sin^3x) =\frac23\sin^3x. ]

The identity method and substitution method therefore agree.

Verification, Common Errors & Study Practice

Verification means differentiating the proposed antiderivative to see whether it returns the original integrand. This check catches missing coefficients, sign errors, and omitted constants. For this problem, the key risk is forgetting the factor (\tfrac13) when integrating (\cos 3x).

Differentiate the result:

[ \frac{d}{dx}\left(\frac12\sin x-\frac16\sin 3x+C\right) =\frac12\cos x-\frac16(3\cos 3x). ]

Simplify:

[ =\frac12\cos x-\frac12\cos 3x =\frac12(\cos x-\cos 3x). ]

The product-to-sum identity showed that (\tfrac12(\cos x-\cos 3x)=\sin x\sin 2x). Differentiating the answer therefore recovers the original integrand.

Common mistakes are usually small but change the answer:

• Dropping the outer (\tfrac12): The identity applies a factor of (\tfrac12) to the whole difference.
• Changing the order inside the difference: The identity uses (\cos(A-B)-\cos(A+B)), not the reverse.
• Forgetting the inner-angle factor: (\int\cos 3x\,dx) is (\tfrac13\sin 3x), not (\sin 3x).
• Leaving out (C): An indefinite integral represents a family of antiderivatives, not just one function.
• Checking only the appearance of the answer: Different-looking forms may be equivalent, so use identities or differentiate.

A practical study routine is to write one line for each decision: identify the identity, substitute the angles, simplify, integrate, and differentiate to check. If the answer is in a different form, compare it with the alternative method or use a trigonometric identity before deciding it is wrong.

Further Applications of Product-to-Sum Techniques

Product-to-sum is useful when an integral contains a product such as (\sin(mx)\sin(nx)), especially when a direct substitution is not clear. The same identity turns two sine factors into cosine terms with sum and difference angles. For a definite integral, evaluate the resulting antiderivative at the upper and lower limits.

Definite integral example

Consider the same integrand from (0) to (\pi):

[ \int_0^\pi\sin x\sin 2x\,dx. ]

Use the antiderivative already found:

[ \left[\frac12\sin x-\frac16\sin 3x\right]_0^\pi. ]

At (x=\pi), both (\sin\pi) and (\sin3\pi) equal zero. At (x=0), both (\sin0) and (\sin0) equal zero. So the definite integral is

[ 0-0=0. ]

This result does not mean the integrand is always zero. The product is positive on part of the interval and negative on another part, so the signed areas cancel. A definite integral measures net signed area, not total area.

When to choose each method

For (\sin x\sin 2x), substitution works neatly after using the double-angle identity. For other products, product-to-sum may be more direct. Neither method is always best; choose the one that turns the integrand into a form you can integrate without adding extra steps.

Extending the identity

For (\sin(mx)\sin(nx)), where (m) and (n) are constants, the same rule gives

[ \sin(mx)\sin(nx) =\frac12\left[\cos((m-n)x)-\cos((m+n)x)\right]. ]

If (m-n) and (m+n) are nonzero, integrate each cosine by dividing by its angle coefficient:

[ \int\sin(mx)\sin(nx)\,dx =\frac12\left[ \frac{\sin((m-n)x)}{m-n} -\frac{\sin((m+n)x)}{m+n} \right]+C. ]

This formula follows from the same steps as the original problem. If one coefficient is zero, handle that term separately because (\int\cos 0\,dx=\int1\,dx=x). The broader lesson is to track the coefficient inside every angle.

Conclusion

The central move is to turn (\sin x\sin 2x) into (\tfrac12(\cos x-\cos 3x)). Integrating then gives (\tfrac12\sin x-\tfrac16\sin 3x+C). Keep the (\tfrac13) factor for the (3x) angle, include (C), and differentiate your result to confirm it.

Frequently Asked Questions

These quick answers review the result, the identity, and the steps most likely to cause confusion. They also distinguish the indefinite integral from a definite one and explain why more than one correct antiderivative form may appear.

What is (\int\sin x\sin 2x\,dx)?
It is (\boxed{\tfrac12\sin x-\tfrac16\sin 3x+C}).

Which identity should I use?
Use (\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]), with (A=x) and (B=2x).

Why does (\cos 3x) integrate to (\tfrac13\sin 3x)?
Because differentiating (\sin 3x) gives (3\cos 3x). Multiplying by (\tfrac13) cancels that extra factor.

Why is the coefficient of (\sin 3x) negative?
The product-to-sum identity subtracts (\cos(A+B)). Here, (A+B=3x), so its antiderivative keeps a negative sign.

Do I need to write (+C)?
Yes, for an indefinite integral. The constant represents all possible vertical shifts of the antiderivative.

Can substitution solve this integral?
Yes. Rewrite (\sin 2x) as (2\sin x\cos x), then use (u=\sin x). This gives (\tfrac23\sin^3x+C), an equivalent answer.

How can I check the answer?
Differentiate it. The derivative simplifies to (\tfrac12(\cos x-\cos 3x)), which equals (\sin x\sin 2x).

Is (\tfrac23\sin^3x+C) a different answer?
No. The triple-angle identity shows that it equals (\tfrac12\sin x-\tfrac16\sin 3x+C).

What is (\int_0^\pi\sin x\sin 2x\,dx)?
It is (0). Evaluating the antiderivative at both endpoints gives zero, because the positive and negative signed areas cancel.

What is the most common error?
Leaving out the factor (\tfrac13) when integrating (\cos 3x). Differentiate the final expression to catch this and other coefficient errors.

(This article was written by one of our staff writers, Alan Westbrook. Visit our Meet the Team page.)