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. :))