Answer
$$\int\frac{1}{t^2}\cos\Big(\frac{1}{t}-1\Big)dt=-\sin\Big(\frac{1}{t}-1\Big)+C$$
Work Step by Step
$$A=\int\frac{1}{t^2}\cos\Big(\frac{1}{t}-1\Big)dt$$
We set $u=\frac{1}{t}-1$
Then $$du=-\frac{1}{t^2}dt$$
That means $$\frac{1}{t^2}dt=-du$$
Therefore, $$A=-\int\cos udu=-\sin u+C$$ $$A=-\sin\Big(\frac{1}{t}-1\Big)+C$$
