(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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(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: $[\sin(t)]_{1}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(1) = 1 - \sin(1)$.

Class 12 Mathematics 2 Integrals: Definite Integrals Definite integration by substitution

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

A
$1$
B
$1 + \sin(1)$
C
$1 - \sin(1)$
D
$\sin(1) - 1$

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2 Integrals: Definite Integrals

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Definite integration by substitution

If $\frac{d[f(x)]}{dx} = g(x)$ for $a \le x \le b$,then $\int_a^b f(x)g(x) dx$ equals

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$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\operatorname{cosec} x \cdot \cot x}{1+\operatorname{cosec}^2 x} d x=$

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If $\int\limits_0^2 375 x^5 (1 + x^2)^{-4} dx = 2^n$,then the value of $n$ is:

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Let $\int_\alpha^{\log _e 4} \frac{dx}{\sqrt{e^{x}-1}}=\frac{\pi}{6}$. Then $e^\alpha$ and $e^{-\alpha}$ are the roots of the equation :

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If $\int_{\log 2}^x \frac{du}{({e^u} - 1)^{1/2}} = \frac{\pi}{6}$,then ${e^x} = $

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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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If $\frac{d[f(x)]}{dx} = g(x)$ for $a \le x \le b$,then $\int_a^b f(x)g(x) dx$ equals

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