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. :))
 A: 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!
A: 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}
A: You need to be careful with your limits of integration. In general, if you're doing a $u$-substitution to solve a definite integral, there are two potential ways to deal with the limits of integration:

(1) Adjust them when you make your substitution.
(2) Don't worry about them initially. Get an antiderivative in terms of $u$, then switch back to the original variable and apply Fundamental Theorem of Calculus.

For an example, let's consider the simple one $\int_2^3xe^{x^2}\,dx$, with the substitution $u=x^2$. In either case, we have $\frac{du}{dx}=2x$, so $du=2x\,dx$, and so $x\,dx=\frac12\,du.$ (Note that I didn't solve for $dx$, here, but rather for $x\,dx$, which are the extraneous $x$-terms when we make the substitution $e^{x^2}\mapsto e^u$ in the original integrand.)
Using method (1), $x=2$ implies $u=2^2=4$, and $x=3$ implies $u=3^2=9$. Hence, $$\int_2^3xe^{x^2}\,dx=\frac12\int_4^9e^u\,du=\frac12\left[e^u\right]_{u=4}^9=\frac12(e^9-e^4).$$
Using method (2), $$\int xe^{x^2}\,dx=\frac12\int e^u\,du=\frac12e^u=\frac12e^{x^2}.$$ (Since we're about to apply the FTC, we don't need an integration constant, here.) Thus, by FTC, we have $$\int xe^{x^2}\,dx=\left[\frac12e^{x^2}\right]_{x=2}^3=\frac12(e^9-e^4).$$
Pick whichever method works for you, and stick with it. (I recommend the first one, personally.) You seem to be favoring the latter. However, you must be cautious with the notation. $\int_0^{\ln(\pi+1)}e^x\sin(e^x-1)\,dx\neq\int_0^{\ln(\pi+1)}\sin u\,du$, for example. As a potential alternative, you could write $$\int_0^{\ln(\pi+1)}e^x\sin(e^x-1)\,dx=\int_{x=0}^{\ln(\pi+1)}\sin u\,du,$$ to make clear that the latter limits of integration are on the variable $x$.
