Question

Use the table of integrals at the back of the text to evaluate the integrals in Exercises 1–26.∫ sin(t / 3) sin(t / 6) dt

Accepted Answer

Recognize that the integral involves the product of two sine functions: $$\sin\left(\frac{t}{3}\right)$$ and $$\sin\left(\frac{t}{6}\right). To $$simplify this, use the product-to-sum identity for sine functions, which states:

$$\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]. $$Apply the product-to-sum formula by letting $$A = \frac{t}{3}$$ and $$B = \frac{t}{6}. $$Substitute these into the identity to rewrite the integrand as:

$$\sin\left(\frac{t}{3}\right) \sin\left(\frac{t}{6}\right) = \frac{1}{2} \left[ \cos\left(\frac{t}{3} - \frac{t}{6}\right) - \cos\left(\frac{t}{3} + \frac{t}{6}\right) \right]. $$Simplify the arguments of the cosine functions inside the brackets:

$$\frac{t}{3} - \frac{t}{6} = \frac{t}{6}$$ and $$\frac{t}{3} + \frac{t}{6} = \frac{t}{2}. So $$the integrand becomes:

$$\frac{1}{2} \left[ \cos\left(\frac{t}{6}\right) - \cos\left(\frac{t}{2}\right) \right]. $$Rewrite the integral using this expression:

$$\int \sin\left(\frac{t}{3}\right) \sin\left(\frac{t}{6}\right) dt = \int \frac{1}{2} \left[ \cos\left(\frac{t}{6}\right) - \cos\left(\frac{t}{2}\right) \right] dt. $$Split the integral into two separate integrals and factor out the constant $$\frac{1}{2}$$:

$$= \frac{1}{2} \int \cos\left(\frac{t}{6}\right) dt - \frac{1}{2} \int \cos\left(\frac{t}{2}\right) dt.

$$Next, use the table of integrals to find the antiderivatives of $$\cos\left(\frac{t}{6}\right)$$ and $$\cos\left(\frac{t}{2}\right)$$, remembering to apply the chain rule in reverse (i.e., multiply by the reciprocal of the inner function's derivative).

Use the table of integrals at the back of the text to evaluate the integrals in Exercises 1–26.
∫ sin(t / 3) sin(t / 6) dt

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Key Concepts

Trigonometric Product-to-Sum Identities

Integration of Basic Trigonometric Functions

Use of Integral Tables