Question

The value of $\int_0^{\pi} \int_0^{\cos y} x \sin y \, dx \, dy$ is given by}

• A. 1/2
• B. 1/3
• C. 1/5
• D. -1/3

Hint:

Since $\cos^2 y$ is an even function and $\sin y$ is symmetric on $[0, \pi]$, we can utilize substitution directly to transform the limits into $[-1, 1]$, making the integration straightforward.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To evaluate a double integral with variable limits, we integrate sequentially from the innermost integral to the outermost integral.

Step 2: Detailed Explanation:
Let the double integral be $I$: \[ I = \int_0^{\pi} \left[ \int_0^{\cos y} x \sin y \, dx \right] dy \]
First, evaluate the inner integral with respect to $x$, treating $y$ as a constant: \[ \int_0^{\cos y} x \sin y \, dx = \sin y \left[ \frac{x^2}{2} \right]_0^{\cos y} = \frac{1}{2} \sin y \cos^2 y \]
Now, substitute this result into the outer integral with respect to $y$: \[ I = \int_0^{\pi} \frac{1}{2} \sin y \cos^2 y \, dy \]
To evaluate this definite integral, we use the method of substitution.
Let: \[ u = \cos y \implies du = -\sin y \, dy \implies \sin y \, dy = -du \]
Change the limits of integration: \[ \text{When } y = 0 \implies u = \cos(0) = 1 \] \[ \text{When } y = \pi \implies u = \cos(\pi) = -1 \]
Substitute these into the integral: \[ I = \frac{1}{2} \int_1^{-1} u^2 (-du) = \frac{1}{2} \int_{-1}^1 u^2 \, du \] \[ = \frac{1}{2} \left[ \frac{u^3}{3} \right]_{-1}^1 = \frac{1}{6} \left[ 1^3 - (-1)^3 \right] = \frac{1}{6} (1 + 1) = \frac{2}{6} = \frac{1}{3} \]

Step 3: Final Answer:
The value of the double integral is 1/3.