How do you find $\int x \cdot {sec}^{2} x$ $int x*sec^2x$ ?

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Answer 1 · 2015-12-11T17:03:07

Integrate by parts (then by subsitution).

$\int x \cdot {sec}^{2} x$ $int x*sec^2x$

Let $u = x$ $u = x$ and $d v = {sec}^{2} x d x$ $dv = sec^2x dx$ .

This makes $d u = d x$ $du=dx$ and $v = tan x$ $v=tanx$

$u v - \int v d u = x tan x - \int tan x d x$ $uv-intvdu = xtanx-int tanx dx$

$= x tan x - \int \frac{sin x}{cos x} d x$ $= xtanx-int sinx/cosx dx$

$= x tan x + ln | cos x | + C$ $= xtanx + ln abs cosx +C$

How do you find ∫x⋅sec2xint x*sec^2x?