0 / 2 cos x - sin x 1 + sin x cos x ? d x =

Question

$\int_{0}^{π / 2} \frac{\cos x - \sin x}{1 + \sin x \cos x} d x =$

Smart Achievers · Accepted Answer

(3)

Ans. 0

Explanation :

$\int_{0}^{π / 2} \frac{\cos x - \sin x}{1 + \sin x \cos x} d x = I$ $\dots (i)$

$Now$ $I = \int_{0}^{π / 2} \frac{\cos (\frac{π}{2} - x) - \sin (\frac{π}{2} - x)}{1 + \sin (\frac{π}{2} - x) \cos (\frac{π}{2} - x)} d x$ $[∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x]$

$= \int_{0}^{π / 2} \frac{\sin x - \cos x}{1 + \sin x \cos x} d x$ $\dots (i i)$

$On adding equation (i) and (ii),$

$2 I = 0 \Rightarrow I = 0$

Solution

Q.
$\int_{0}^{π / 2} \frac{\cos x - \sin x}{1 + \sin x \cos x} d x =$

1 2

2 - 2

3 0

4 None of these

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Solution

Q. ∫0π/2cosx-sinx1+sinxcosx dx=

1 2

2 - 2

3 0

4 None of these

Ans. (3) Ans. 0 Explanation : ∫0π/2cosx-sinx1+sinxcosx dx=I ⋯(i) Now I=∫0π/2cos(π2-x)-sin(π2-x)1+sin(π2-x)cos(π2-x) dx [∵∫0af(x) dx=∫0af(a-x) dx] =∫0π/2sinx-cosx1+sinx cosx dx ⋯(ii) On adding equation (i) and (ii), 2I=0 ⇒ I=0

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Q.
$\int_{0}^{π / 2} \frac{\cos x - \sin x}{1 + \sin x \cos x} d x =$