Question

Integration of x. Sec square x into dx

Answer 1

Answer:

which takes the form ∫udv=uv−∫vdu.

For the given integral ∫xsec2(x)dx, we want to choose a value of u that gets simpler when we differentiate it and a value of dv that is easily integrated.

So, let:

{u=x ⇒ du=dxdv=sec2(x)dx ⇒ v=tan(x)

We then have:

∫xsec2(x)dx=uv−∫vdu

∫xsec2(x)dx=xtan(x)−∫tan(x)dx

You may have the integral of tan(x) memorized. If not, it's easy to find:

∫xsec2(x)dx=xtan(x)−∫sin(x)cos(x)dx

Let t=cos(x), implying that dt=−sin(x)dx:

∫xsec2(x)dx=xtan(x)+∫−sin(x)cos(x)dx

∫xsec2(x)dx=xtan(x)+∫1tdt

This is a common integral:

∫xsec2(x)dx=xtan(x)+ln(|t|)+C

Working back from t=cos(x):

∫xsec2(x)dx=xtan(x)+ln(|cos(x)|)+C

here is your ans..