Integrate
They gave me the U for this integral but im not sure how im suppose to work it into the problem. Can someone show me the steps to do this?
$\displaystyle \int\frac{\sec^2\left(\frac{1}{x^{39}}\right)}{x^{ 40}}dx$
If we let $\displaystyle z=\frac{1}{x^{39}}$ then $\displaystyle dz=\frac{-1}{x^{40}}$. So
$\displaystyle \int\frac{\sec^2\left(\frac{1}{x^{39}}\right)}{x^{ 40}}dx\overbrace{\mapsto}^{z=\frac{1}{x^{39}}}-\int\sec^2(z)dz$