What is the value of the definite integral \int_{0}^{\frac{\pi}{3}} \cos(2x) \, dx? — JAMB UTME

What is the value of the definite integral $\int_{0}^{\frac{\pi}{3}} \cos(2x) \, dx$ ?

\frac{\sqrt{3}}{4}

The antiderivative of

\cos(2x)

\frac{1}{2}\sin(2x)

. Substituting the limits gives

\frac{1}{2}\sin\left(\frac{2\pi}{3}\right) - \frac{1}{2}\sin(0) = \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) - 0 = \frac{\sqrt{3}}{4}

, which is correct.

Find the antiderivative of

\cos(2x)

\int \cos(2x) \, dx = \frac{1}{2}\sin(2x) + C

Using standard trigonometric integration rules,

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

Substitute the upper limit

x = \frac{\pi}{3}

and lower limit

x = 0

\left[\frac{1}{2}\sin(2x)\right]_{0}^{\frac{\pi}{3}} = \frac{1}{2}\sin\left(\frac{2\pi}{3}\right) - \frac{1}{2}\sin(0)

Apply the Fundamental Theorem of Calculus:

\int_{a}^{b} f(x)dx = F(b) - F(a)

Evaluate the trigonometric values and simplify

\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) - 0 = \frac{\sqrt{3}}{4}

\sin\left(\frac{2\pi}{3}\right) = \sin(120^\circ) = \frac{\sqrt{3}}{2}

and

\sin(0) = 0

Definite Integration of Trigonometric Functions

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