<s>[INST] Definite integrals using u substitution (verification needed) Would someone mind verifying this?
$
\int_{0}^{\ln(\pi + 1)}e^x \sin(e^x - 1) \space dx
$
$
u = e^x - 1
\Rightarrow \frac{du}{dx} = e^x
\Rightarrow du = e^x \space dx
\Rightarrow dx = \frac{1}{e^x} \space du
$
$
\int_{0}^{\ln(\pi + 1)} e^x \sin(u) \frac{1}{e^x} \space du
= \int_{0}^{\ln(\pi + 1)} \sin(u) \space du
= -\cos(u) \space |_{0}^{\ln(\pi + 1)}
= -\cos(e^x - 1) \space |_{0}^{\ln(\pi + 1)}
= [-\cos(e^{\ln(\pi + 1)} - 1)] - [-\cos(e^{0} - 1)]
= [-\cos(\pi + 1-1)] - [-\cos(1 - 1)]
= [-\cos(\pi)] - [-\cos(0)]
= -\cos(\pi) + 1
$
and a second one
$\int_{\frac{\pi^2}{4}}^{\pi^2}\frac{\cos(\sqrt{x})}{\sqrt(x)} \space dx \\
$
$
u = \sqrt{x}
\Rightarrow \frac{du}{dx} = \frac{1}{2 \sqrt{x}}
\Rightarrow du = \frac{1}{2 \sqrt{x}} \space dx
\Rightarrow dx = \frac{du}{\frac{1}{2 \sqrt{x}}} = 2\sqrt{x} \space du
$
$
\int_{\frac{\pi^2}{4}}^{\pi^2} = \frac{\cos(u)}{\sqrt(x)} 2\sqrt{x} \space du
= 2 \int_{\frac{\pi^2}{4}}^{\pi^2} \cos(u) \space du
= 2\sin(\sqrt{x}) \space |_{\frac{\pi^2}{4}}^{\pi^2}
$
$
= 2\sin(\sqrt{x}) \space |_{\frac{\pi^2}{4}}^{\pi^2}
= [2\sin(\sqrt{\pi^2})] - [2\sin(\sqrt{\frac{\pi^2}{4}})]
= [2\sin(\pi)]-[2\sin\frac{\pi}{2}]
= [2(0)]-[2(1)]
= 0 - 2
= -2
$
(Thanks. :))
 [/INST] When you were doing the u subsitution, why did you bring the e^x to the du side? When subsituting, you want to eliminate all functions of x and change them into a function in terms of u. Also you will get a new interval when making u-substitutions!
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