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

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(C) આપેલ છે કે $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$. $\sin 2x = 2 \sin x \cos x$ નો ઉપયોગ કરતા: $I(x) = \int e^{\sin^2 x} (2 \sin x

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) આપેલ છે કે $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$. $\sin 2x = 2 \sin x \cos x$ નો ઉપયોગ કરતા: $I(x) = \int e^{\sin^2 x} (2 \sin x \cos^2 x - \sin x) dx$. અહીં,$\frac{d}{dx} (e^{\sin^2 x} \cos x) = e^{\sin^2 x} (2 \sin x \cos x) \cos x + e^{\sin^2 x} (-\sin x) = e^{\sin^2 x} (2 \sin x \cos^2 x - \sin x) = e^{\sin^2 x} (\cos x \sin 2x - \sin x)$ થાય છે. તેથી,$I(x) = e^{\sin^2 x} \cos x + C$. $I(0) = 1$ આપેલ હોવાથી,$1 = e^{\sin^2 0} \cos 0 + C \Rightarrow 1 = 1 \cdot 1 + C \Rightarrow C = 0$. આમ,$I(x) = e^{\sin^2 x} \cos x$. તેથી $I\left(\frac{\pi}{3}\right) = e^{\sin^2(\pi/3)} \cos(\pi/3) = e^{3/4} \cdot \frac{1}{2} = \frac{1}{2} e^{\frac{3}{4}}$.

જો $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$ અને $I(0) = 1$ હોય,તો $I\left(\frac{\pi}{3}\right)$ ની કિંમત શોધો.

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