If and , then is equal to:

Explanation: I = ∫ e s i n 2 x ( cos x sin 2 x − sin x ) d x I = ∫ II e s i n 2 x sin 2 x I cos x d x − ∫ e s i n 2 x sin x d x I = cos x ∫ e s i n 2 x sin 2 x d x − ∫ (( − sin x ) ⋅ ∫ e s i n 2 x sin 2 x d x ) d x − ∫ e s i n 2 x sin x d x + C Let t = sin 2 x ⇒ d t = 2 sin x cos x d x = sin 2 x d x So I = cos x ∫ e t d t + ∫ ( sin x ∫ e t d t ) d x − ∫ e s i n 2 x sin x d x + C ⇒ I = cos x e t + ∫ sin x e t d x − ∫ e s i n 2 x sin x d x + C ⇒ I = cos x e s i n 2 x + ∫ sin x e s i n 2 x d x − ∫ e s i n 2 x sin x d x + C ⇒ I = cos x ⋅ e s i n 2 x + C but I ( 0 ) = 1 ⇒ 1 = cos 0 × e s i n 2 0 + C ⇒ 1 = 1 × e 0 + C ⇒ C = 0 so I = cos x

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$ and $I(0) = 1$ , then $I\left(\frac{\pi}{3}\right)$ is equal to:

Choose the correct answer:

Explanation

$I = \int e s i n 2 x (cos x sin 2 x - sin x) d x$

$I = \int II e s i n 2 x sin 2 x I cos x d x - \int e s i n 2 x sin x d x$

$I = cos x \int e s i n 2 x sin 2 x d x - \int ((- sin x) \cdot \int e s i n 2 x sin 2 x d x) d x - \int e s i n 2 x sin x d x + C$

Let $t = sin 2 x$ $\Rightarrow d t = 2 sin x cos x d x = sin 2 x d x$

So $I = cos x \int e t d t + \int (sin x \int e t d t) d x - \int e s i n 2 x sin x d x + C$

$\Rightarrow I = cos x e t + \int sin x e t d x - \int e s i n 2 x sin x d x + C$

$\Rightarrow I = cos x e s i n 2 x + \int sin x e s i n 2 x d x - \int e s i n 2 x sin x d x + C$

$\Rightarrow I = cos x \cdot e s i n 2 x + C$

but $I (0) = 1$ $\Rightarrow 1 = cos 0 \times e s i n 2 0 + C$ $\Rightarrow 1 = 1 \times e 0 + C \Rightarrow C = 0$

so $I = cos x \cdot e s i n 2 x$

and $I (3 π) = cos 3 π e s i n 2 (3 π)$

$I (3 π) = 21 \times e 43$

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