Question

The value of the integral $\int_{0}^{\pi / 2} \sqrt{\cos x \sin 2x} dx$ is equal to

• A. $\frac{\sqrt{2}}{3}$
• B. $\frac{2\sqrt{2}}{3}$
• C. $\frac{2}{3}$
• D. $\frac{2}{3\sqrt{2}}$
• E. $\frac{3\sqrt{2}}{2}$

Hint:

Always look for the derivative of a part of the function appearing as a factor in the integrand. Here, $\cos x$ was perfectly placed to be the $du$ for $u = \sin x$ once the square root was simplified.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We evaluate a definite integral of a trigonometric function. We can simplify the integrand using the double-angle identity to prepare it for substitution.
Step 2: Key Formula or Approach:
Use \(\sin 2x = 2\sin x \cos x\).
Set \(u = \sin x\), finding \(du = \cos x dx\).
Step 3: Detailed Explanation:
Simplify the expression under the square root:
\[ I = \int_{0}^{\pi/2} \sqrt{\cos x (2\sin x \cos x)} dx \] \[ I = \int_{0}^{\pi/2} \sqrt{2\cos^2 x \sin x} dx \] Pull out constants and perfect squares:
Since \(\cos x \geq 0\) on \([0, \pi/2]\), \(\sqrt{\cos^2 x} = \cos x\).
\[ I = \sqrt{2} \int_{0}^{\pi/2} \cos x \sqrt{\sin x} dx \] Use u-substitution. Let \(u = \sin x\).
Then \(du = \cos x dx\).
Change the bounds:
When \(x = 0, u = \sin(0) = 0\).
When \(x = \pi/2, u = \sin(\pi/2) = 1\).
Substitute into the integral:
\[ I = \sqrt{2} \int_{0}^{1} \sqrt{u} \, du = \sqrt{2} \int_{0}^{1} u^{1/2} du \] Integrate:
\[ I = \sqrt{2} \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{1} = \sqrt{2} \left( \frac{2}{3}(1)^{3/2} - 0 \right) \] \[ I = \frac{2\sqrt{2}}{3} \] Step 4: Final Answer:
The value is \(\frac{2\sqrt{2}}{3}\).