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Let $\frac{d}{d x} F(x)=\left(\frac{e^{\text {sinx }}}{x}\right) x>0$. If $\int_1^4 \frac{3}{x} e^{\text {sin }^3} d x=F(k)-F(1)$ then one of the possible values of $k$, is
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Verified Answer
The correct answer is:
64
64
$$
\frac{d}{d x} F(x)=\frac{e^{\sin x}}{x} \text { or } \int_1^4 \frac{3}{x} e^{\sin x^3} d x=\int_1^4 \frac{3 x^2}{x^3} e^{\sin x^3} d x
$$
Let $\mathrm{x}^3=\mathrm{t}, 3 \mathrm{x}^2 \mathrm{dx}=\mathrm{dt}$
when $x=1, t=1 \& x-4, t=64$
$$
\begin{aligned}
& \mathrm{F}(\mathrm{t})=\int_1^{64} \frac{\mathrm{e}^{\sin t}}{\mathrm{t}} \mathrm{dt}=\int_1^{64} \mathrm{~F}(\mathrm{t}) \mathrm{dt}=\mathrm{F}(64)-\mathrm{F}(1) \\
& \mathrm{K}=64
\end{aligned}
$$
\frac{d}{d x} F(x)=\frac{e^{\sin x}}{x} \text { or } \int_1^4 \frac{3}{x} e^{\sin x^3} d x=\int_1^4 \frac{3 x^2}{x^3} e^{\sin x^3} d x
$$
Let $\mathrm{x}^3=\mathrm{t}, 3 \mathrm{x}^2 \mathrm{dx}=\mathrm{dt}$
when $x=1, t=1 \& x-4, t=64$
$$
\begin{aligned}
& \mathrm{F}(\mathrm{t})=\int_1^{64} \frac{\mathrm{e}^{\sin t}}{\mathrm{t}} \mathrm{dt}=\int_1^{64} \mathrm{~F}(\mathrm{t}) \mathrm{dt}=\mathrm{F}(64)-\mathrm{F}(1) \\
& \mathrm{K}=64
\end{aligned}
$$
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