Answer
$1$
Work Step by Step
In order to integrate this function, we have to use an «u» substitution.
Choose an u, derivate it:
$ u = \cos x +1$
$ du = -\sin x dx$
$ -du = \sin x dx$
Change limits:
$x_1 =\pi/2$
$u_1= \cos (\pi/2) +1=1$
$x_2 =\pi$
$u_2= \cos (\pi) +1=0$
Then substitute:
\[ = -\int_{1}^{0} 6u^5 du = \int_{0}^{1} 6u^5 du\]
Then integrate
\[ =u^6 \bigg ]_{0}^{1} \]
Apply calculus 2nd theorem and reduce:
\[ =(1)^6-(0)^6=1 \]
