Answer
$a.\quad[\sinh^{-1}u]_{0}^{0}=0$
$b.\quad[\ln(x+\sqrt{x^{2}+1})]_{0}^{0}=0$
Work Step by Step
$(a)$
Substitute:$\quad \left[\begin{array}{ll}
u=\sin x & du=\cos xdx\\
x=0 & \rightarrow u=0\\
x=\pi & \rightarrow u=0
\end{array}\right]$
$I=\displaystyle \int_{0}^{0}\frac{1}{\sqrt{1+u^{2}}}du$
Using the table "lntegrals leading to inverse hyperbolic functions"
$\displaystyle \int\frac{du}{\sqrt{a^{2}+u^{2}}}=\sinh^{-1}(\frac{u}{a})+C$,
we would proceed solving in terms of hyperbolic functions, but, since the bounds of our new definite integral are equal, we know that the integral equals 0.
Just to demonstrate, $(a=1)$
$I=[\sinh^{-1}u]_{0}^{0}=\sinh^{-1}0-\sinh^{-1}0$
$=0-0$
$=0$
$(b)$
Again, we know that the definite integral equals zero,
but to illustrate the conversion of inverse hyperbolic to ln, we would use the formulas in the box above these exercises, by which
$\sinh^{-1}x=\ln(x+\sqrt{x^{2}+1}),$
So,
$I=[\ln(x+\sqrt{x^{2}+1})]_{0}^{0}$
$=\ln(0+\sqrt{0+1})-\ln(0+\sqrt{0+1})$
$=\ln 1-\ln 1$
$=0-0$
$=0$
