Target Exam

CUET

Subject

Section B1

Chapter

Definite Integration

Question:

$\int\limits_0^{\frac{\pi}{2}} \frac{1-\cot x}{cosec~ x+\cos x} d x=$

Options:

0

$\frac{\pi}{4}$

$\infty$

$\frac{\pi}{12}$

Correct Answer:

0

Explanation:

The correct answer is Option (1) → 0

$I = \int\limits_{0}^{\pi/2} \frac{1 - \cot x}{\text{cosec}\, x + \cos x} \, dx$

$= \int\limits_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx$

Let

$I = \int\limits_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx$

Substitute $x \to \frac{\pi}{2} - x$:

$I = \int\limits_{0}^{\pi/2} \frac{\cos x - \sin x}{1 + \sin x \cos x} \, dx = -I$

$\Rightarrow I = -I \Rightarrow 2I = 0 \Rightarrow I = 0$