Aspire Faculty ID #11953 · Topic: NIMCET 2025 · Just now
NIMCET 2025

The value of $\frac{d}{dx}\int ^{2\sin x}_{\sin {x}^2}{e}^{{t}^2}dt$ at $x=\pi$

Solution

We want to find $\dfrac{d}{dx}\left(\int_{\sin^2 x}^{2\sin x} e^{t^2}, dt\right)$ at $x=\pi$. 
Using Leibniz rule: $\dfrac{d}{dx}\left(\int_{a(x)}^{b(x)} f(t) dt\right) = f(b(x)) b'(x) ;-; f(a(x)) a'(x)$ 

Here $a(x)=\sin^2 x$, $b(x)=2\sin x$, $f(t)=e^{t^2}$. 
Compute derivatives:
$a'(x)=2\sin x\cos x$ 
$b'(x)=2\cos x$ 
So the derivative is: 
$e^{(2\sin x)^2}(2\cos x)-e^{(\sin^2 x)^2}(2\sin x\cos x)$ 

Now evaluate at $x=\pi$
$\sin\pi=0,\quad \cos\pi=-1$ 

Thus: $b'(\pi)=2(-1)=-2$ 
$a'(\pi)=0$ 

Therefore: $e^{0}(-2)-e^{0}(0)=-2$ 
So the final answer is: $-2$
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