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 simplify the trigonometric expression inside the square root using the double-angle identity for sine and then use the method of substitution.
Key Formula or Approach:
Identity: $\sin 2x = 2 \sin x \cos x$.
Substitution: Let $u = \sin x$, then $du = \cos x dx$.

Step 2: Detailed Explanation:
1. Simplify the integrand:
\[ I = \int_{0}^{\pi / 2} \sqrt{\cos x (2 \sin x \cos x)} dx \]
\[ I = \int_{0}^{\pi / 2} \sqrt{2 \sin x \cos^2 x} dx \]
\[ I = \sqrt{2} \int_{0}^{\pi / 2} \cos x \sqrt{\sin x} dx \]
2. Use substitution:
Let $u = \sin x \implies du = \cos x dx$.
When $x = 0$, $u = \sin 0 = 0$.
When $x = \pi/2$, $u = \sin(\pi/2) = 1$.
3. Compute the new integral:
\[ I = \sqrt{2} \int_{0}^{1} \sqrt{u} du = \sqrt{2} \int_{0}^{1} u^{1/2} du \]
\[ I = \sqrt{2} \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{1} = \sqrt{2} \cdot \frac{2}{3} [1 - 0] \]
\[ I = \frac{2\sqrt{2}}{3} \]

Step 3: Final Answer:
The value of the integral is $\frac{2\sqrt{2}}{3}$.