Question

Evaluate the definite integral:

$\int_{0}^{\pi/2} \sin x \cos 2x \, dx$

• A. -1/3
• B. 1/3
• C. -2/3
• D. 2/3

Answer 1

Answer: (A)

-1/3

Answer 2

To evaluate the integral $\int_{0}^{\pi/2} \sin x \cos 2x \, dx$, we can use the product-to-sum trigonometric identity:

$\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$

In our case, $A = x$ and $B = 2x$, so:

$\sin x \cos 2x = \frac{1}{2} [\sin(x+2x) + \sin(x-2x)] = \frac{1}{2} [\sin(3x) - \sin(x)]$

Now, we can rewrite the integral as:

$\int_{0}^{\pi/2} \sin x \cos 2x \, dx = \frac{1}{2} \int_{0}^{\pi/2} [\sin(3x) - \sin(x)] \, dx$

We can split the integral into two parts:

$\frac{1}{2} \left[ \int_{0}^{\pi/2} \sin(3x) \, dx - \int_{0}^{\pi/2} \sin(x) \, dx \right]$

Now, we evaluate each integral separately:

1. $\int_{0}^{\pi/2} \sin(3x) \, dx = \left[ -\frac{1}{3} \cos(3x) \right]_{0}^{\pi/2} = -\frac{1}{3} \cos\left(\frac{3\pi}{2}\right) - \left(-\frac{1}{3} \cos(0)\right) = -\frac{1}{3}(0) + \frac{1}{3}(1) = \frac{1}{3}$

2. $\int_{0}^{\pi/2} \sin(x) \, dx = \left[ -\cos(x) \right]_{0}^{\pi/2} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = -0 + 1 = 1$

Substitute these results back into the original expression:

$\frac{1}{2} \left[ \frac{1}{3} - 1 \right] = \frac{1}{2} \left[ -\frac{2}{3} \right] = -\frac{1}{3}$

Therefore, the value of the definite integral is $-\frac{1}{3}$.