What's the integral of $\int {(tan (x))}^{2} \cdot sec (x) d x$ $int (tan(x))^2 * sec(x) dx$ ?

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Answer 1 · 2016-08-02T19:06:44

$\frac{sec x tan x - ln (| sec x + tan x |)}{2} + C$ $(secxtanx-ln(abs(secx+tanx)))/2+C$

We have:

$I = \int {tan}^{2} x sec x d x$ $I=inttan^2xsecxdx$

Write ${tan}^{2} x$ $tan^2x$ as ${sec}^{2} x - 1$ $sec^2x-1$ .

$I = \int ({sec}^{2} x - 1) sec x d x$ $I=int(sec^2x-1)secxdx$

$I = \int {sec}^{3} x d x - \int sec x d x$ $I=intsec^3xdx-intsecxdx$

The second is a commonly known integral:

$I = \int {sec}^{3} x d x - ln (| sec x + tan x |)$ $I=intsec^3xdx-ln(abs(secx+tanx))$

Now, for the remaining integral, we will try to use integration by parts, which takes the form:

$\int u d v = u v - \int v d u$ $intudv=uv-intvdu$

So, let:

${(u=secx" "=>" "du=secxtanxdx),(dv=sec^2xdx" "=>" "v=tanx):}$

Thus:

$I=secxtanx-intsecxtan^2xdx-ln(abs(secx+tanx))$

Now, notice that we have $intsecxtan^2xdx$ in the problem here, which is what we started with. So, we also know that:

$intsecxtan^2xdx=secxtanx-intsecxtan^2xdx-ln(abs(secx+tanx))$

Add $intsecxtan^2xdx$ to both sides:

$2intsecxtan^2xdx=secxtanx-ln(abs(secx+tanx))$

Divide both sides by $2$ :

$intsecxtan^2xdx=(secxtanx-ln(abs(secx+tanx)))/2+C$

What's the integral of int (tan(x))^2 * sec(x) dx∫(tan(x))2⋅sec(x)dxint (tan(x))^2 * sec(x) dx?