Question

Let $\frac{d}{dx}F\left(x\right)=\left(\frac{e^{\sin x}}{x}\right)$, where $x>0$. If $\underset{1}{\overset{4}{\int }}\frac{3}{x}{e}^{\sin x^3}dx=F\left(k\right)-F\left(1\right)$, then the possible value of $k$ is

Answer 1

For $\underset{1}{\overset{4}{\int }}\frac{3}{x}{e}^{\sin x^3}dx$
Putting $x^3=t\Rightarrow 3x^{2}dx=dt$
When $x=1\Rightarrow t=1;x=4\Rightarrow t=64$
Now,
$\underset{1}{\overset{4}{\int }}\frac{3}{x}{e}^{\sin x^3}dx=\underset{1}{\overset{64}{\int }}\frac{e^{\sin t}}{t}dt=\underset{1}{\overset{64}{\int }}\frac{d}{dt}F\left(t\right)\cdot dt=F\left(64\right)-F\left(1\right)\therefore k=64$