-0-2 cox x-sin x-1-sin x cos xdx -

∫_0^π/2 cox x sin x/1+sin x cos xdx =I ⋯⋯ (i) Now I= ∫_0^π/2 cos ( π/2 x ) sin ( π/2 x )/1+sin ( π/2 x )cos ( π/2 x )dx = ∫_0^π/2 sin x cos x/1+si ...

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Byju's Answer

Standard XIII

Mathematics

Integration of Piecewise Continuous Functions

∫0π/2 cox x-s...

Question

$\int_{0}^{\frac{π}{2}} \frac{c o x x - s i n x}{1 + s i n x c o s x} d x =$

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-2

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Solution

The correct option is C \N
$\int_{0}^{\frac{π}{2}} \frac{c o x x - s i n x}{1 + s i n x c o s x} d x = I \dots \dots (i)$
$N o w I = \int_{0}^{\frac{π}{2}} \frac{c o s (\frac{π}{2} - x) - s i n (\frac{π}{2} - x)}{1 + s i n (\frac{π}{2} - x) c o s (\frac{π}{2} - x)} d x$
$= \int_{0}^{\frac{π}{2}} \frac{s i n x - c o s x}{1 + s i n x c o s x} \dots \dots (i i)$
$O n a d d i n g, 2 I = 0 \Rightarrow I = 0$

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∫π20 cox x−sin x1+sin x cos xdx=

The correct option is C \N∫π20 cox x−sin x1+sin x cos xdx=I ⋯⋯(i) Now I=∫π20 cos(π2−x)−sin(π2−x)1+sin(π2−x)cos(π2−x)dx =∫π20 sin x−cos x1+sin x cos x ⋯⋯(ii) On adding, 2I=0⇒I=0

$\int_{0}^{\frac{π}{2}} \frac{c o x x - s i n x}{1 + s i n x c o s x} d x =$