Introduction

The value of \( \int_0^{\pi/2} x \sin x \, dx \) is a classic definite integration problem frequently asked in JEE, board exams, and competitive tests, and it is best solved using the integration by parts technique.

Mastering this question strengthens understanding of how to combine algebraic and trigonometric functions inside definite integrals with proper application of limits.

Step-by-Step Solution

1. Recall Integration by Parts Formula

For two functions \( u(x) \) and \( v(x) \),

\[ \int u \, dv = uv – \int v \, du \]

This formula is used when the integrand is a product, such as \( x \sin x \).

2. Choose \( u \) and \( dv \)

Take:

\[ u = x \implies du = dx \]
\[ dv = \sin x \, dx \implies v = -\cos x \]

These choices follow the ILATE rule (Inverse, Log, Algebraic, Trigonometric, Exponential), where the algebraic function \( x \) is chosen as \( u \).

3. Apply to Indefinite Integral

Using the formula,

\[ \int x \sin x \, dx = x(-\cos x) – \int (-\cos x) \, dx \]

which simplifies to

\[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx \]

Now integrate \( \cos x \):

\[ \int \cos x \, dx = \sin x \]

So the antiderivative becomes

\[ \int x \sin x \, dx = -x \cos x + \sin x + C \]

4. Evaluate Definite Integral from 0 to \( \pi/2 \)

Evaluate

\[ \int_0^{\pi/2} x \sin x \, dx = \left[ -x \cos x + \sin x \right]_0^{\pi/2} \]

Upper limit \( x = \pi/2 \):

\[ \cos(\pi/2) = 0, \quad \sin(\pi/2) = 1 \]
\[ -(\pi/2) \cdot 0 + 1 = 1 \]

Lower limit \( x = 0 \):

\[ \cos 0 = 1, \quad \sin 0 = 0 \]
\[ -0 \cdot 1 + 0 = 0 \]

5. Final Value

Subtract lower-limit value from upper-limit value:

\[ \int_0^{\pi/2} x \sin x \, dx = 1 – 0 = 1 \]

So, the value of \( \int_0^{\pi/2} x \sin x \, dx \) is 1.

Explanation of MCQ Options

In many exam settings, this question appears as a multiple-choice problem with options like:

• A) 0: Incorrect, because \( x \sin x \) is positive on \( (0, \pi/2) \), so the definite integral must be positive.
• B) \( \pi/2 \): Incorrect; this value corresponds to integrals like \( \int_0^\pi \sin 2x \, dx \).
• C) 1: Correct; detailed integration by parts shows the exact value \( \int_0^{\pi/2} x \sin x \, dx = 1 \).
• D) \( \pi^2/4 \): Incorrect; this appears in problems like \( \int_0^\pi x \sin 2x \, dx \).