Evaluate $$\int_{0.1}^{1}|(\pi)( x^{-2})sin(\pi \cdot x^{-1})|dx$$
The above has to be computed without a calculator.
I know that $$\frac{d}{dx}[cos(\pi \cdot x^{-1})] = (\pi)( x^{-2})sin(\pi \cdot x^{-1})$$
Applying the limits to the left hand side of the above equation, the initial integral evaluates to $-2$. However, the $18$. How do I get 18?
As you have already pointed out, substituting $u=\frac{\pi}{x}$ gives:
$$\int\frac{\pi}{x^2}sin\ \frac{\pi}{x}\ dx=\int-sin\ u\ du=cos\ u\ +C=cos\ \frac{\pi}{x}\ +C$$
The rest of the problem is dealing with the absolute value. Note that $x$ is strictly positive.
$$\int_{0.1}^1|\frac{\pi}{x^2}sin\ \frac{\pi}{x}|dx=\int_{0.1}^1\frac{\pi}{x^2}|sin\ \frac{\pi}{x}|dx$$
Using the same substitution, and putting bounds in terms of $u$:
$$=\int_\pi^{10\pi}|cos\ u|du$$
There are a number of ways of evaluating this intergal. One is to break in into subintervals where $cos\ u$ is strictly positive or negative and applying sign changes accordingly. Alternately, you can use geometry, and the observation that the integral between two consecutive roots of the cosine function is $\pm 2$.
$$=|\int_{\pi}^{\frac{3\pi}{2}}cos\ u\ du|+|\int_{\frac{3\pi}{2}}^{\frac{5\pi}{2}}cos\ u\ du|+...+|\int_{\frac{17\pi}{2}}^{\frac{19\pi}{2}}cos\ u\ du|+|\int_{\frac{19\pi}{2}}^{10\pi}cos\ u\ du|$$
$$=1+2+...+2+1=1+8\cdot 2+1=18$$
