The value of the definite integral,intlimits_ | vedclass

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Question

(C) Let $t = e^{x^2}$.Then,$dt = e^{x^2} \cdot 2x dx$.When $x = 0$,$t = e^{0^2} = e^0 = 1$.When $x = \sqrt{\ln(\frac{\pi}{2})}$,$t = e^{(\sqrt{\ln(\fr

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$1 - \sin(1)$

Vedclass · Answer

(C) Let $t = e^{x^2}$.Then,$dt = e^{x^2} \cdot 2x dx$.When $x = 0$,$t = e^{0^2} = e^0 = 1$.When $x = \sqrt{\ln(\frac{\pi}{2})}$,$t = e^{(\sqrt{\ln(\frac{\pi}{2})})^2} = e^{\ln(\frac{\pi}{2})} = \frac{\pi}{2}$.The integral becomes $\int_{1}^{\frac{\pi}{2}} \cos(t) dt$.Evaluating the integral: $[\sin(t)]_{1}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(1) = 1 - \sin(1)$.

The value of the definite integral,$\int\limits_0^{\sqrt {\ln \left( {\frac{\pi }{2}} \right)} } {\cos \left( {{e^{{x^2}}}} \right)} \cdot 2x {e^{{x^2}}}dx$ is

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