यदि I(x) = int e^{sin^2 x} (cos x sin 2x - si | vedclass

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(C) दिया गया है $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$. $\sin 2x = 2 \sin x \cos x$ का उपयोग करने पर: $I(x) = \int e^{\sin^2 x} (2 \s

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

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(C) दिया गया है $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$. $\sin 2x = 2 \sin x \cos x$ का उपयोग करने पर: $I(x) = \int e^{\sin^2 x} (2 \sin x \cos^2 x - \sin x) dx$. यहाँ,$\frac{d}{dx} (e^{\sin^2 x} \cos x) = e^{\sin^2 x} (2 \sin x \cos x) \cos x + e^{\sin^2 x} (-\sin x) = e^{\sin^2 x} (2 \sin x \cos^2 x - \sin x) = e^{\sin^2 x} (\cos x \sin 2x - \sin x)$ होता है। अतः,$I(x) = e^{\sin^2 x} \cos x + C$. चूँकि $I(0) = 1$,इसलिए $1 = e^{\sin^2 0} \cos 0 + C \Rightarrow 1 = 1 \cdot 1 + C \Rightarrow C = 0$. अतः,$I(x) = e^{\sin^2 x} \cos x$. इसलिए $I\left(\frac{\pi}{3}\right) = e^{\sin^2(\pi/3)} \cos(\pi/3) = e^{3/4} \cdot \frac{1}{2} = \frac{1}{2} e^{\frac{3}{4}}$.

यदि $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$ और $I(0) = 1$ है,तो $I\left(\frac{\pi}{3}\right)$ का मान ज्ञात कीजिए।

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