Question

The value of \(\displaystyle\int^{\frac{\pi}{16}}_{0}\cos6x\cos2xdx\) is equal to

• A. \(\frac{1+\sqrt 2}{16}\)
• B. \(\frac{1+\sqrt 2}{8}\)
• C. \(\frac{2+\sqrt 2}{16}\)
• D. \(\frac{-1+\sqrt 2}{16}\)
• E. \(\frac{-1+\sqrt 2}{8}\)

Answer 1

The Correct Option is A

We are asked to evaluate the integral \[ I = \int_0^{\frac{\pi}{16}} \cos 6x \cos 2x \, dx. \] We will use the product-to-sum identity for cosines: \[ \cos A \cos B = \frac{1}{2} \left( \cos(A - B) + \cos(A + B) \right). \] Here, \( A = 6x \) and \( B = 2x \), so applying the identity: \[ \cos 6x \cos 2x = \frac{1}{2} \left( \cos(6x - 2x) + \cos(6x + 2x) \right) = \frac{1}{2} \left( \cos 4x + \cos 8x \right). \] Thus, the integral becomes: \[ I = \int_0^{\frac{\pi}{16}} \frac{1}{2} \left( \cos 4x + \cos 8x \right) \, dx. \] Now, integrate each term: \[ I = \frac{1}{2} \left[ \int_0^{\frac{\pi}{16}} \cos 4x \, dx + \int_0^{\frac{\pi}{16}} \cos 8x \, dx \right]. \] The integral of \( \cos 4x \) is: \[ \int \cos 4x \, dx = \frac{1}{4} \sin 4x, \] and the integral of \( \cos 8x \) is: \[ \int \cos 8x \, dx = \frac{1}{8} \sin 8x. \] Evaluating both integrals from 0 to \( \frac{\pi}{16} \), we get: \[ I = \frac{1}{2} \left[ \frac{1}{4} \left( \sin \left( 4 \times \frac{\pi}{16} \right) - \sin 0 \right) + \frac{1}{8} \left( \sin \left( 8 \times \frac{\pi}{16} \right) - \sin 0 \right) \right]. \] Simplifying: \[ I = \frac{1}{2} \left[ \frac{1}{4} \sin \frac{\pi}{4} + \frac{1}{8} \sin \frac{\pi}{2} \right]. \] Now, \( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \) and \( \sin \frac{\pi}{2} = 1 \), so: \[ I = \frac{1}{2} \left[ \frac{1}{4} \times \frac{\sqrt{2}}{2} + \frac{1}{8} \times 1 \right] = \frac{1}{2} \left[ \frac{\sqrt{2}}{8} + \frac{1}{8} \right] = \frac{1}{16} \left( 1 + \sqrt{2} \right). \]

The correct option is (A) : \(\frac{1+\sqrt 2}{16}\)

Answer 2

We want to evaluate the definite integral: \[\int_{0}^{\frac{\pi}{16}} \cos(6x) \cos(2x) \, dx\]

We can use the product-to-sum trigonometric identity: \[\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)]\] In our case, \(A = 6x\) and \(B = 2x\), so \[\cos(6x) \cos(2x) = \frac{1}{2} [\cos(6x - 2x) + \cos(6x + 2x)] = \frac{1}{2} [\cos(4x) + \cos(8x)]\]

Therefore, the integral becomes: \[\int_{0}^{\frac{\pi}{16}} \cos(6x) \cos(2x) \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{16}} [\cos(4x) + \cos(8x)] \, dx\]

We integrate each term separately: \[\int \cos(4x) \, dx = \frac{1}{4} \sin(4x) + C\] \[\int \cos(8x) \, dx = \frac{1}{8} \sin(8x) + C\]

So, \[\frac{1}{2} \int_{0}^{\frac{\pi}{16}} [\cos(4x) + \cos(8x)] \, dx = \frac{1}{2} \left[\frac{1}{4} \sin(4x) + \frac{1}{8} \sin(8x)\right]_{0}^{\frac{\pi}{16}}\]

Evaluate at the limits of integration: \[\frac{1}{2} \left[\left(\frac{1}{4} \sin\left(4\cdot\frac{\pi}{16}\right) + \frac{1}{8} \sin\left(8\cdot\frac{\pi}{16}\right)\right) - \left(\frac{1}{4} \sin(0) + \frac{1}{8} \sin(0)\right)\right]\] \[= \frac{1}{2} \left[\left(\frac{1}{4} \sin\left(\frac{\pi}{4}\right) + \frac{1}{8} \sin\left(\frac{\pi}{2}\right)\right) - (0 + 0)\right]\] \[= \frac{1}{2} \left[\frac{1}{4} \cdot \frac{\sqrt{2}}{2} + \frac{1}{8} \cdot 1\right] = \frac{1}{2} \left[\frac{\sqrt{2}}{8} + \frac{1}{8}\right] = \frac{1}{2} \cdot \frac{\sqrt{2} + 1}{8} = \frac{1 + \sqrt{2}}{16}\]

Therefore, the definite integral is: \[\int_{0}^{\frac{\pi}{16}} \cos(6x) \cos(2x) \, dx = \frac{1 + \sqrt{2}}{16}\]