JEE MAIN - Mathematics (2023 - 10th April Morning Shift - No. 1)
If $$I(x) = \int {{e^{{{\sin }^2}x}}(\cos x\sin 2x - \sin x)dx} $$ and $$I(0) = 1$$, then $$I\left( {{\pi \over 3}} \right)$$ is equal to :
$$ - {e^{{3 \over 4}}}$$
$$ - {1 \over 2}{e^{{3 \over 4}}}$$
$${e^{{3 \over 4}}}$$
$${1 \over 2}{e^{{3 \over 4}}}$$
Explanation
$$
\begin{aligned}
& \text { Given, } I(x)=\int e^{\sin ^2 x}(\cos x \sin 2 x-\sin x) d x \\\\
& =\int e^{\sin ^2 x} \cdot \cos x \cdot \sin 2 x d x-\int \sin x e^{\sin ^2 x} d x \\\\
& =\int \frac{\cos x}{\mathrm{I}} \cdot \frac{e^{\sin ^2 x} \cdot \sin 2 x}{\mathrm{II}} d x-\int \sin x \cdot e^{\sin ^2 x} d x \\\\
& =\cos x \cdot e^{\sin ^2 x}-\int(-\sin x) e^{\sin ^2 x} d x-\int \sin x e^{\sin ^2 x} d x \\\\
& =\cos x \cdot e^{\sin ^2 x}+\int \sin x e^{\sin ^2 x} \cdot d x-\int \sin x \cdot e^{\sin ^2 x} d x+C
\end{aligned}
$$
$I(x)=e^{\sin ^2 x} \cdot \cos x+C$
Given, $I(0)=1 \Rightarrow C=0$
So, $ I(x)=e^{\sin ^2 x} \cdot \cos x$
$$ \Rightarrow I(\pi / 3)=e^{3 / 4} \cdot \frac{1}{2}=\frac{e^{3 / 4}}{2} $$
$I(x)=e^{\sin ^2 x} \cdot \cos x+C$
Given, $I(0)=1 \Rightarrow C=0$
So, $ I(x)=e^{\sin ^2 x} \cdot \cos x$
$$ \Rightarrow I(\pi / 3)=e^{3 / 4} \cdot \frac{1}{2}=\frac{e^{3 / 4}}{2} $$
Comments (0)
