If I ( x ) = e sin 2 x ( cos x sin 2 x - sin x ) ? d x and I ( 0 ) = 1 , then I ( 3 ) is equal to [2023]

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Solution

Q.
$If I (x) = \int e^{\sin^{2} x} (\cos x \sin 2 x - \sin x) d x and I (0) = 1,$ $then I (\frac{π}{3}) is equal to$ [2023]

1 $- e^{\frac{3}{4}}$

2 $e^{\frac{3}{4}}$

3 $\frac{1}{2} e^{\frac{3}{4}}$

4 $- \frac{1}{2} e^{\frac{3}{4}}$

Ans.
(3)

$I (x) = \int e^{\sin^{2} x} (\cos x \sin 2 x - \sin x) d x$

$= \int e^{g (x)} [f (x) g' (x) + f' (x)] d x$ $[where g (x) = \sin^{2} x and f (x) = \cos x]$

$= e^{g (x)} f (x) + C = e^{\sin^{2} x} \cdot \cos x + C$

$As I (0) = 1,$

$∴$ $e^{\sin^{2} 0} \cos (0) + C = 1$ $\Rightarrow C = 0$ $∴$ $I (x) = e^{\sin^{2} x} \cos x$

$Hence, I (\frac{π}{3}) = e^{\sin^{2} \frac{π}{3}} \cos \frac{π}{3} = \frac{1}{2} e^{\frac{3}{4}}$

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