-0-2-0-2sinx-y dxdy is-

I=∫_0^π/2∫_0^π/2sin(x+y) dxdy =∫_0^π/2[ cos(x+y)]_0^π /2 dy =∫_0^π/2 [ cos ( π/2+y )+cos y ]dy =∫_0^π/2[sin y +cosy] dy =[ cos y+sin y]_0^π /2 = cosπ/2+sinπ/ ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Inverse Trigonometric Functions

∫0π/2∫0π/2sin...

Question

$\int_{0}^{\frac{π}{2}} \int_{0}^{\frac{π}{2}} s i n (x + y) d x d y$ is:

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π

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\frac{π}{2}

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2

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Solution

The correct option is D $2$
$I = \int_{0}^{\frac{π}{2}} \int_{0}^{\frac{π}{2}} s i n (x + y) d x d y$
$= \int_{0}^{\frac{π}{2}} [- c o s (x + y)]_{0}^{π / 2} d y$
$= \int_{0}^{\frac{π}{2}} [- c o s (\frac{π}{2} + y) + c o s y] d y$
$= \int_{0}^{\frac{π}{2}} [s i n y + c o s y] d y$
$= [- c o s y + s i n y]_{0}^{π / 2}$
$= - c o s \frac{π}{2} + s i n \frac{π}{2} - (- c o s 0 + s i n 0)$
$I = - 0 + 1 - (- 1 + 0)$
$I = 2$

Suggest Corrections

∫π20∫π20sin(x+y)dxdy is:

The correct option is D 2I=∫π20∫π20sin(x+y)dxdy =∫π20[−cos(x+y)]π/20dy =∫π20[−cos(π2+y)+cosy]dy =∫π20[siny+cosy]dy =[−cosy+siny]π/20 =−cosπ2+sinπ2−(−cos0+sin0) I=−0+1−(−1+0) I=2

$\int_{0}^{\frac{π}{2}} \int_{0}^{\frac{π}{2}} s i n (x + y) d x d y$ is: