Indefinite Integral: $\int{x\cos{(x+1)}}dx$ I am trying to solve the following indefinite integral:
$$\int{x\cos{(x+1)}}dx$$
But I keep running into problems as I am thinking of solving it using parts in which I did the following:
$$
u = x+1 \\
du = 1\ dx \\
\int{x\cos{(x+1)}}dx = \int{x\cos{u}}du
$$
Then from $u = x+1$ I get $x = u-1$ then:
$$\int{x\cos{u}}du = \int{(u-1)\cos{u}}du$$
Which appears that I am doing something really wrong. How do I proceed?

Your substitution is correct: the easiest way of proceeding with your method is to notice that  $\int{(u-1)\cos{u}}\;du$ can be divided into two integrals
$$\int{(u-1)\cos{u}}\;du=\int u\cos{u}\;du-\int{\cos{u}}\;du$$
The first integral of the two can be found by integrating by parts
$$\int u\cos{u}\;du=u\sin u-\int \sin u\;du$$
Therefore using your method proceeds as
$$\int{(u-1)\cos{u}}\;du=u\sin u-\int \sin u\;du-\int{\cos{u}}\;du$$
which becomes
$$\int{(u-1)\cos{u}}\;du=u\sin u+ \cos u-\sin{u}+C$$
