What is the value of the definite integral \int_{0}^{\frac{\pi}{6}} \cos(3x) \, dx? — JAMB UTME

What is the value of the definite integral $\int_{0}^{\frac{\pi}{6}} \cos(3x) \, dx$ ?

The value of the definite integral is

\frac{1}{3}

The antiderivative of

\cos(3x)

\frac{1}{3}\sin(3x)

. Substituting the upper boundary

x = \frac{\pi}{6}

yields

\frac{1}{3}\sin\left(\frac{\pi}{2}\right) = \frac{1}{3}

, and substituting the lower boundary

x = 0

yields

0

. Subtracting the lower bound evaluation from the upper bound evaluation gives

\frac{1}{3}

Find the indefinite integral of

\cos(3x)

\int \cos(3x) \, dx = \frac{1}{3}\sin(3x) + C

Using the standard integration rule

\int \cos(kx) \, dx = \frac{1}{k}\sin(kx) + C

Apply the upper limit of integration

x = \frac{\pi}{6}

\frac{1}{3}\sin\left(3 \cdot \frac{\pi}{6}\right) = \frac{1}{3}\sin\left(\frac{\pi}{2}\right) = \frac{1}{3}(1) = \frac{1}{3}

Substituting the upper limit into the antiderivative.

Apply the lower limit of integration

x = 0

and subtract

\frac{1}{3} - \frac{1}{3}\sin(0) = \frac{1}{3} - 0 = \frac{1}{3}

Evaluating the antiderivative at the limits according to the Fundamental Theorem of Calculus.

Definite Integration of Trigonometric Functions

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