Q.If $f(x)=\displaystyle\int_{1}^{x}\dfrac{e^{\sin u}}{u}\,du$, $x>1$ and $\displaystyle\int_{1}^{3}\dfrac{e^{\sin x^2}}{x}\,dx=\dfrac12\left[f(a)-f(1)\right]$, then one of the possible value of a is :

The derivative of a variable-upper-limit integral is simply the integrand evaluated at the upper limit, by the Fundamental Theorem of Calculus.

1. $f(x)=\displaystyle\int_0^xt\cos t\,dt$, where the integrand is $g(t)=t\cos t$ (a continuous function of $t$). …

The substitution $t=x^2$ converts the given integral directly into $\frac12[f(9)-f(1)]$, identifying $a=9$.

1. In $\displaystyle\int_1^3\dfrac{e^{\sin x^2}}{x}\,dx$, let $t=x^2$, so $dt=2x\,dx\Rightarrow\dfrac{dx}x=\dfrac{dt}{2t}$ (since $\dfrac1x\,dx=\dfrac{dt}{2x^2}=\dfrac{dt}{2t}$).
2. Limits: $x=1\Rightarrow t=1$; $x=3\Rightarrow t=9$.
3. So $\displaystyle\int_1^3\dfrac{e^{\sin x^2}}{x}\,dx=\int_1^9\dfrac{e^{\sin t}}{2t}\,dt=\dfrac12\int_1^9\dfrac{e^{\sin t}}{t}\,dt$. …

Rewriting the integrand in the standard $\int dx/\sqrt{a^2-x^2}$ form (with $a=2/3$ after factoring out $3$) integrates directly to an inverse sine, evaluated at the given limits.

1. Factor $9$ out of the square root: $4-9x^2=9\left(\dfrac49-x^2\right)=9\left(\left(\dfrac23\right)^2-x^2\right)$, so $\sqrt{4-9x^2}=3\sqrt{\left(\dfrac23\right)^2-x^2}$.
2. So $\displaystyle\int_0^{2/3}\dfrac{dx}{\sqrt{4-9x^2}}=\dfrac13\int_0^{2/3}\dfrac{dx}{\sqrt{(2/3)^2-x^2}}$. …

Using the standard antiderivative $\int\tan x\,dx=-\ln|\cos x|+C$ and evaluating between the given limits.

1. $\displaystyle\int\tan x\,dx=\int\dfrac{\sin x}{\cos x}\,dx=-\ln|\cos x|+C$.
2. So $\displaystyle\int_0^{\pi/3}\tan x\,dx=\Big[-\ln|\cos x|\Big]_0^{\pi/3}=-\ln\left(\cos\dfrac{\pi}{3}\right)+\ln(\cos0)$. …

Rewriting the integrand in the standard $\int dx/\sqrt{a^2-x^2}$ form and evaluating the limits gives $\dfrac{\pi}{6}$.

1. We need $\displaystyle\int_0^{2/3}\dfrac{dx}{\sqrt{4-9x^2}}$.
2. Factor out $9$ from inside the square root: $4-9x^2=9\left(\dfrac49-x^2\right)$, so $\sqrt{4-9x^2}=3\sqrt{\left(\dfrac23\right)^2-x^2}$.
3. So the integral becomes $\displaystyle\int_0^{2/3}\dfrac{dx}{3\sqrt{(2/3)^2-x^2}}=\dfrac13\int_0^{2/3}\dfrac{dx}{\sqrt{(2/3)^2-x^2}}$.
4. Use the standard result $\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\!\left(\dfrac{x}{a}\right)+C$ with $a=\dfrac23$: $\dfrac13\left[\sin^{-1}\!\left(\dfrac{x}{2/3}\right)\right]_0^{2/3}=\dfrac13\left[\sin^{-1}\!\left(\dfrac{3x}{2}\right)\right]_0^{2/3}$. …