How do you evaluate the integral $\int x {tan}^{2} x$ $int xtan^2x$ ?

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How do you evaluate the integral $\int x {tan}^{2} x$ $int xtan^2x$ ?

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Answer 1 · 2017-03-09T07:12:12

I would begin by using the identity ${tan}^{2} (θ) = {sec}^{2} (θ) - 1$ $tan^2(theta)=sec^2(theta)-1$ .

$\Rightarrow \int x ({sec}^{2} (x) - 1) d x$ $=>intx(sec^2(x)-1)dx$

$\int x {sec}^{2} (x) d x - \int x d x$ $intxsec^2(x)dx-intxdx$

The RH is a basic integral. Let's continue with the left. Now we will use integration by parts.

$u = x s p a c e d v = {sec}^{2} (x) d x$ $u=x color(white)(space) dv=sec^2(x)dx$

$d u = d x s p a c e v = tan (x)$ $du=dxcolor(white)(space)v=tan(x)$

Our integral now takes the form of $u v - \int v d u$ $uv-intvdu$ :

$x tan (x) - \int tan (x) d x$ $xtan(x)-inttan(x)dx$

The RH is a basic integral.

Combining this with what we found above:

$x tan (x) - \int tan (x) d x - \int x d x$ $xtan(x)-inttan(x)dx-intxdx$

Integrating, we get:

$\Rightarrow x tan (x) + ln (| cos (x) |) - \frac{x^{2}}{2} + C$ $=>xtan(x)+ln(|cos(x)|)-x^2/2+C$

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How do you evaluate the integral $\int x {tan}^{2} x$ $int xtan^2x$ ?

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