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MystifiedBoy @MystifiedBoy

October 2022 1 38 Report

Hey Brainiacs ! Find derivative
[tex]\clubsuit \: \boxed{ \displaystyle \rm y=\int_{0}^{ lnx} sine^t\ dt}[/tex]

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EngrJohann

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$\Large \mathbb{SOLUTION:}$

$\begin{array}{l} \displaystyle y=\int_{0}^{\ln x} \sin (e^t)\ dt \\ \\ \begin{aligned} \textsf{Let }u &= e^t \\ du &= e^t dt \longrightarrow dt = \dfrac{du}{u} \end{aligned} \\ \\ \textsf{At }x = 0, \: u = e^0 = 1 \\ \textsf{At }x = \ln x, \: u = e^{\ln x} = x \\ \\ \displaystyle \implies y=\int_{1}^{x} \dfrac{\sin u}{u} du \\ \\ \textsf{Now we need to find the derivative of }y. \\ \\ \textsf{By Second Fundamental Theorem of Calculus,} \\ \\ \displaystyle \dfrac{d}{dx} \int_{a}^{x} f(t)\ dt = f(x), \textsf{where }a\textsf{ is constant.} \\ \\ \displaystyle \implies \dfrac{dy}{dx} = \dfrac{d}{dx} \int_{1}^{x} \dfrac{\sin u}{u}\ du \\ \\ \quad \therefore \boxed{\dfrac{dy}{dx} = \dfrac{\sin x}{x}} \\ \\ \\ \textsf{Or we can just skip the substitution part. Take} \\ \textsf{the derivative of the upper limit which is }\ln x\\ \textsf{then multiply by }\sin (e^{\ln x}) = \sin(x). \\ \\ \qquad \dfrac{d}{dx}\ln x = \dfrac{1}{x} \\ \\ \displaystyle \therefore y' = \dfrac{d}{dx} \int_{0}^{\ln x} \sin (e^t)\ dt = \boxed{\dfrac{\sin x}{x}} \end{array}$

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EngrJohann Thank you

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