Definite Integration
Definite Integration
nta_abhyas_2025
Grade 12

Question:

The value of the integral $I = \int_0^\pi [\sin x + \cos x](\cos x - \sin x) dx$ is equal to (where $[.]$ denotes the greatest integer function)
$\sqrt{2}$
$2\sqrt{2}$
1
$\sqrt{2} - 1$

Step-by-Step Solution

Key Concept: Clever substitution $\sin x + \cos x = t$ transforms the denominator into a recognizable form
Let $\sin x + \cos x = t$, so $(\cos x - \sin x)dx = dt$. The integral becomes $I = \int_1^{\sqrt{2}} \frac{1}{t^2 - 1} dt = \sqrt{2} - 1$. This uses the substitution technique and standard integral formulas for rational functions.
Correct Answer: $\sqrt{2} - 1$

Master Definite Integration with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free