# Somewhat easy integral

So this is my first year of calculus and we're doing integrals using u-substitution

I'm having trouble with this integral:

integral of (1/x^2)*sec(1/x)*tan(1/x)dx

i let u=(1/x)
so du=(1/x^2)dx
so du(x^2)=dx

If I do this, the integral works out perfectly since the x^2 cancel out

so it equals sec(u) + C

HOWEVER, my question is
if i let u = (1/x)

why don't i replace the (1/x^2) (in the beginning of the integral) with u^2

CAN SOMEONE HELP ME?!?!

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First of all, the derivative of $$\frac{1}{x}$$ is not $$\frac{1}{x^2}$$

Second of all, if $$u^2 = \frac{1}{x^2}$$

then $$2udu = \frac{-2}{x^3}dx$$

And you haven't made anything simpler for yourself.

the derivative of 1/x is -1/x^2 correct? sorrry

but if i were to do a u substitution like

integral of tan^4(x)*sec^2(x)

and i let u=tan x
then du*sec^2x=dx

and that leads to the integral of u^4, correct??? i'm confused

I don't follow. The derivative of u with respect to x is du / dx and the derivative of tan x is (secx)^2. so du / dx = (secx)^2.

right. so dx = du / (secx)^2 so
if i substitute that back into the original integral the (secx)^2 cancel out.
is the clearer?

But you can't let u = tan(x) because there is no tan(x) in the integral. There's a tan(1/x).

The difference is that you are saying "let u^2 = some function, whereas in this question, we let u = some function, and when you substitute u back into the integral, it just happens to become u^3.

What you want to do is let u = some function so that du = (some function that's already inside the integral)dx. This simplifies things in terms of u and du with no X's.