<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] You arrived at the correct results in a rather unusual way. Here is how it should be:
$u = e^x- 1\Rightarrow \frac{du}{dx} = e^x\Rightarrow du = e^x \space dx$
and $0\le x\le\ln(\pi + 1)\Rightarrow 0\le u\le \pi$
Therefore,
\begin{gather}
\int_{0}^{\ln(\pi + 1)} e^x \sin(e^x-1) \space dx
= \int_{0}^{\pi} \sin(u) \space du
= -\cos(u) \space |_{0}^{\pi}
= [-\cos(\pi)] - [-\cos(0)]
= 1+1=2
\end{gather}
</s>