What is the value of $\int_{0}^{\pi} x |\cos x| \sin x \, dx$?
Show Hint
Whenever the term $x$ is multiplied by a symmetric trigonometric expression over the interval $[0, \pi]$, King's property always simplifies the integral by converting $\int_0^\pi x f(x) \, dx$ to $\frac{\pi}{2} \int_0^\pi f(x) \, dx$.
This is a standard shortcut for competitive exams.
Step 1 : Understanding the Question:
The question asks us to evaluate the definite integral $I = \int_{0}^{\pi} x |\cos x| \sin x \, dx$. Step 2 : Key Formulas and Approach:
We will use the properties of definite integrals, specifically King's Property:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
Applying this property allows us to eliminate the linear factor $x$ from the integrand. Step 3 : Detailed Explanation:
Let the given integral be:
\[ I = \int_{0}^{\pi} x |\cos x| \sin x \, dx \quad \text{--- (Equation 1)} \]
Using King's Property, replace $x$ with $\pi - x$:
\[ I = \int_{0}^{\pi} (\pi - x) |\cos(\pi - x)| \sin(\pi - x) \, dx \]
We know that:
• $\sin(\pi - x) = \sin x$
• $\cos(\pi - x) = -\cos x \implies |\cos(\pi - x)| = |-\cos x| = |\cos x|$
Substitute these back into the integral:
\[ I = \int_{0}^{\pi} (\pi - x) |\cos x| \sin x \, dx \]
\[ I = \pi \int_{0}^{\pi} |\cos x| \sin x \, dx - \int_{0}^{\pi} x |\cos x| \sin x \, dx \]
Notice that the second integral is the original integral $I$:
\[ I = \pi \int_{0}^{\pi} |\cos x| \sin x \, dx - I \]
\[ 2I = \pi \int_{0}^{\pi} |\cos x| \sin x \, dx \]
\[ I = \frac{\pi}{2} \int_{0}^{\pi} |\cos x| \sin x \, dx \]
Let us evaluate the remaining integral $J = \int_{0}^{\pi} |\cos x| \sin x \, dx$.
Since the integrand $|\cos x| \sin x$ is symmetric about $\frac{\pi}{2}$ (the function value at $\pi - x$ is equal to the function value at $x$), we can write:
\[ J = 2 \int_{0}^{\pi/2} \cos x \sin x \, dx \]
Using the double-angle identity $2 \sin x \cos x = \sin 2x$:
\[ J = \int_{0}^{\pi/2} \sin 2x \, dx \]
\[ J = \left[ -\frac{\cos 2x}{2} \right]_{0}^{\pi/2} \]
\[ J = -\frac{1}{2} \left( \cos \pi - \cos 0 \right) \]
\[ J = -\frac{1}{2} (-1 - 1) = 1 \]
Now substitute $J = 1$ back into the expression for $I$:
\[ I = \frac{\pi}{2} \cdot (1) = \frac{\pi}{2} \] Step 4 : Final Answer:
The value of the definite integral is $\frac{\pi}{2}$.
This corresponds to Option (A).
