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Question Number 28894 by amit96 last updated on 01/Feb/18

Commented by abdo imad last updated on 01/Feb/18

$l e t p u t f (x) = F (x, y) = \int_{0}^{x^{2} + y^{2}} {c o s}^{2} (t + x) d t w e h a v e b y c h .$ $t + x = u \Rightarrow f (x) = \int_{x}^{x^{2} + y^{2} + x} {c o s}^{2} u d u = \int_{x}^{x^{2} + y^{2} + x} \frac{1 + c o s (2 u)}{2} d u$ $= \frac{1}{2} (x^{2} + y^{2}) + {[\frac{1}{4} s i n (2 u)]}_{x}^{x^{2} + y^{2} + x}$ $= \frac{1}{2} (x^{2} + y^{2}) + \frac{1}{4} (s i n (2 x^{2} + 2 y^{2} + 2 x) - s i n (2 x))$ $\frac{\partial F}{\partial x} (x, y) = x + \frac{1}{4} ((4 x + 2) c o s (2 x^{2} + 2 y^{2} + 2 x) - 2 c o s (2 x))$ $\frac{\partial F}{\partial x} (0, y) = \frac{1}{4} (2 c o s (2 y^{2}) - 2) = \frac{1}{2} c o s (2 y^{2}) - \frac{1}{2} a n d a n y a n s w e r$ $g i v e n i s c o r r e c t .!$
Commented by abdo imad last updated on 01/Feb/18

$\frac{\partial F}{\partial x} (0, y) = \frac{1}{2} (1 - 2 {s i n}^{2} (2 y^{2})) - \frac{1}{2} = - {s i n}^{2} (2 y^{2}) .$
Commented by mrW2 last updated on 01/Feb/18

$- {s i n}^{2} (2 y^{2}) o r - {s i n}^{2} (y^{2}) ?$
Commented by amit96 last updated on 01/Feb/18

$y o u a r e r i g h t a t m r w 2$
Commented by mrW2 last updated on 01/Feb/18

$s i r, y o u h a d t w o d i f f e r e n t s o l u t i o n s :$ $\frac{\partial F}{\partial x} (0, y) == \frac{1}{2} c o s (2 y^{2}) - \frac{1}{2} . . . . (I)$ $a n d$ $\frac{\partial F}{\partial x} (0, y) = - \sin^{2} (2 y^{2}) . . . . (I I)$ $i f (I) i s c o r r e c t, t h e n - \sin^{2} (y^{2}) i s$ $c o r r e c t, n o t - \sin^{2} (2 y^{2}) . s o I a s k e d$ $y o u w h i c h i s c o r r e c t : - \sin^{2} (y^{2}) o r$ $- \sin^{2} (2 y^{2}) ?$
	Answered by mrW2 last updated on 01/Feb/18

	$\int \cos^{2} t d t = \frac{1}{2} \int (1 + \cos 2 t) d t = \frac{t}{2} + \frac{\sin 2 t}{4} + C$ $F (x, y) = \int_{0}^{x^{2} + y^{2}} \cos^{2} (t + x) d t$ $= {[\frac{t + x}{2} + \frac{\sin 2 (t + x)}{4}]}_{0}^{x^{2} + y^{2}}$ $= [\frac{x^{2} + y^{2} + x}{2} - \frac{x}{2} + \frac{\sin 2 (x^{2} + y^{2} + x)}{4} - \frac{\sin 2 x}{4}]$ $= [\frac{x^{2} + y^{2}}{2} + \frac{\sin 2 (x^{2} + y^{2} + x)}{4} - \frac{\sin 2 x}{4}]$ $\frac{\partial F}{\partial x} = x + \frac{\cos 2 (x^{2} + y^{2} + x) \times 2 (2 x + 1)}{4} - \frac{\cos 2 x}{4} \times 2$ $\frac{\partial F}{\partial x} = x + \frac{(2 x + 1) \cos 2 (x^{2} + y^{2} + x)}{2} - \frac{\cos 2 x}{2}$ $\frac{\partial F}{\partial x} (0, y) = \frac{\cos 2 (y^{2})}{2} - \frac{1}{2} = \frac{1 - 2 \sin^{2} y^{2}}{2} - \frac{1}{2} = - \sin^{2} y^{2}$ $\Rightarrow n o n e o f t h e a n s w e r s i s c o r r e c t .$
	Commented by amit96 last updated on 01/Feb/18

	$t h a n k s$
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