What is the Integral of ${tan}^{2} (x) sec (x)$ $tan^2(x)sec(x)$ ?

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What is the Integral of ${tan}^{2} (x) sec (x)$ $tan^2(x)sec(x)$ ?

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Answer 1 · 2017-02-03T23:13:05

$\frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln (| sec (x) + tan (x) |) + C$ $1/2sec(x)tan(x)-1/2ln(abs(sec(x)+tan(x)))+C$

Explanation:

$I = \int {tan}^{2} (x) sec (x) d x$ $I=inttan^2(x)sec(x)dx$

Let ${tan}^{2} (x) = {sec}^{2} (x) - 1$ $tan^2(x)=sec^2(x)-1$ which comes from the Pythagorean identity:

$I = \int ({sec}^{2} (x) - 1) sec (x) d x = \int {sec}^{3} (x) d x - \int sec (x) d x$ $I=int(sec^2(x)-1)sec(x)dx=intsec^3(x)dx-intsec(x)dx$

The integral of $sec (x)$ $sec(x)$ is well known:

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

The integral of ${sec}^{3} (x)$ $sec^3(x)$ can be found through integration by parts with $u = sec (x)$ $u=sec(x)$ and $d v = {sec}^{2} (x) d x$ $dv=sec^2(x)dx$ at this link. You can also see how to integrate $sec (x)$ $sec(x)$ there.

$I = (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln (| sec (x) + tan (x) |)) - ln (| sec (x) + tan (x) |)$ $I=(1/2sec(x)tan(x)+1/2ln(abs(sec(x)+tan(x))))-ln(abs(sec(x)+tan(x)))$

$I = \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln (| sec (x) + tan (x) |) + C$ $I=1/2sec(x)tan(x)-1/2ln(abs(sec(x)+tan(x)))+C$

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What is the Integral of ${tan}^{2} (x) sec (x)$ $tan^2(x)sec(x)$ ?

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